# Inequality 57 (George Basdekis)

Problem:

If $x_1,x_2,\ldots, x_n$ are non-negative real numbers with sum equal to $1$ then find the maximum value of the expression

$\displaystyle P(x_1,x_2,\ldots,x_n)=x_1^4x_2^2+x_2^4x_3^2+\cdots+x_n^4x_1^2+n^{3(n-1)}x_1^4x_2^4\cdots x_n^4.$

Solution:

First we may, without loss of generality, assume that $x_1=\max\{x_1,x_2,\ldots,x_n\}.$ Now, we shall prove that

$\displaystyle P(x_1,x_2,\ldots,x_n)\leq P(x_1,x_2+x_3+\cdots+x_n,0,0,\ldots,0).$

It holds that

\displaystyle \begin{aligned} P(x_1&,x_2+x_3+\cdots+x_n,0,0,\ldots,0)=x_1^4(x_2+x_3+\cdots+x_n)^2\geq\\&\geq 2x_1^4(x_2x_3+x_3x_4+\cdots+x_nx_1)+x_1^4(x_2^2+x_n^2)\\&\geq (x_1^4x_2x_3+x_1^4x_3x_4+\cdots+x_1^4x_nx_1)+(x_2^4x_3^2+\cdots+x_{n-1}^4x_n^2)+x_1^4x_2^2+x_n^4x_1^2\\&\geq x_1^4x_2x_3+(x_1^4x_2^2+x_1^4x_3^2+x_2^4x_3^2+\cdots+x_n^4x_1^2).\end{aligned}

Therefore, we only need to prove that

$\displaystyle x_1^4x_2x_3\geq n^{3(n-1)}x_1^4x_2^4\cdots x_n^4.$

For $n=3$ we get

$\displaystyle x_1^4x_2x_3\geq 2^6x_1^4x_2^4x_3^4$

or equivalently the above Inequality reduces to the $\displaystyle x_2^3x_3^3\leq\frac{1}{2^6}.$ But this one holds due to the AM-GM Inequality since we have

$\displaystyle x_2^3x_3^3\leq\left[\frac{x_2+x_3}{2}\right]^6\leq\left[\frac{x_1+x_2+x_3}{2}\right]^6=\frac{1}{2^6}.$

Now, for $n>3$ on the other side, we only need to show that

$\displaystyle n^{3(n-1)}x_1^4x_2^4\cdots x_n^4\leq x_1^4x_2x_3$

or

$\displaystyle x_2^3x_3^3x_4^4\cdots x_n^4\leq\frac{1}{n^{3(n-1)}}.$

But, again, from the AM-GM Inequality we acquire

\displaystyle \begin{aligned}x_2^3x_3^3x_4^4\cdots x_n^4\leq (x_2x_3\cdots x_n)^3&\leq\left[\frac{x_2+x_3+\cdots+x_n}{n-1}\right]^{3(n-1)}\\&\leq\left[\frac{x_1+x_2+\cdots+x_n}{n-1}\right]^{3(n-1)}\\&=\frac{1}{n^{3(n-1)}}.\end{aligned}

So, finally, we are left to find the maximum value of the expression

$\displaystyle P(x_1,x_2+x_3+\cdots+x_n,0,0,\ldots,0)=x_1^4(1-x_1)^2.$

We have according to the AM-GM Inequality

\displaystyle \begin{aligned} x_1^4(1-x_1)^2&=4^4\cdot 2^2\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{1-x_1}{2}\cdot\frac{1-x_1}{2}\\&\leq 4^4\cdot 2^2\cdot\left[\frac{4\cdot\frac{x_1}{4}+2\cdot\frac{1-x_1}{2}}{4+2}\right]^6\\&=\frac{16}{729}.\end{aligned}

Thus, the maximum value of the expression $P$ is equal to $16/729$ and it is obtained if and only if $x_1=2/3,\, x_2=1/3,\, x_3=x_4=\cdots=x_n=0$ Q.E.D.

# Inequality 56 (George Basdekis)

Problem:

Let $a,b,c$ be positive real numbers. Prove the Inequality

$\displaystyle \frac{a\sqrt{a}}{2a+b}+\frac{b\sqrt{b}}{2b+c}+\frac{c\sqrt{c}}{2c+a}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt{3}\sqrt[3]{abc}}.$

Solution:

First, we observe that from the AM-GM Inequality we have that

$\displaystyle 3a(a+2b)\leq\left[\frac{4a+2b}{2}\right]^2=(2a+b)^2$

or equivalently

$\displaystyle \frac{3a}{(2a+b)^2}\leq\frac{1}{a+2b}\Leftrightarrow \frac{3a^3}{(2a+b)^2}\leq\frac{a^2}{a+2b}$

and now taking square roots for both sides, we see that

$\displaystyle \sqrt{3}\cdot\frac{a\sqrt{a}}{2a+b}\leq\frac{a}{\sqrt{a+2b}}.$

Therefore, we acquire

$\displaystyle \sqrt{3}\sum_{cyc}\frac{a\sqrt{a}}{2a+b}\leq\sum_{cyc}\frac{a}{\sqrt{a+2b}}.$

Now, from the Cauchy-Schwarz Inequality we have for the right hand side

\displaystyle \begin{aligned}\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2&\leq\sum_{cyc}[a(a+2c)]\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\\&=(a+b+c)^2\sum_{cyc}\frac{a}{(a+2b)(a+2c)}.\end{aligned}

Moreover, once again, from the AM-GM Inequality we have for the denominators of the above sum that

$\displaystyle (a+2b)(a+2c)\geq 9\sqrt[3]{a^2b^2c^2}$

and thus, it will hold that

\displaystyle \begin{aligned}(a+b+c)^2\sum_{cyc}\frac{a}{(a+2b)(a+2c)}&\leq (a+b+c)^2\sum_{cyc}\frac{a}{9\sqrt[3]{a^2b^2c^2}}\\&=\frac{(a+b+c)^3}{9\sqrt[3]{a^2b^2c^2}}.\end{aligned}

Finally, we have proved that

$\displaystyle \left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2\leq\frac{(a+b+c)^3}{9\sqrt[3]{a^2b^2c^2}}\Leftrightarrow \sum_{cyc}\frac{a}{\sqrt{a+2b}}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt[3]{abc}}.$

The Inequality we have proved is

$\displaystyle \sqrt{3}\cdot\sum_{cyc}\frac{a\sqrt{a}}{2a+b}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt[3]{abc}}$

which also is the Inequality we are given to prove. The proof is completed Q.E.D.

# Inequality 55 (Unknown Author)

Problem:

If $x,y$ are positive real numbers then prove that

$\displaystyle x^y+y^x>1.$

Solution:

The Inequality is obvious for $x,y\geq 1$. So, let’s consider the case where $x,y\in (0,1)$. We will now divide the Inequality into two cases.

• Case 1st: $x^{y-1}>y^{x-1}.$ Then it holds that

$\displaystyle (x^y+y^x)^{1/x}=\left[y^x\left(\frac{x^y}{y^x}+1\right)\right]^{1/x}=y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}$

therefore, according to Bernulli’s Inequality it will hold that

\displaystyle \begin{aligned}y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}&\geq y\left(1+\frac{x^{y-1}}{y^x}\right)\\&\geq y\left(1+\frac{y^{x-1}}{y^x}\right)\\&=y+1>1.\end{aligned}

• Case 2nd: $y^{x-1}>x^{y-1}.$ Then it is

$\displaystyle (x^y+y^x)^{1/y}=\left[x^y\left(\frac{y^x}{x^y}+1\right)\right]^{1/y}=x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}$

therefore, once again from Bernulli’s Inequality we get

\displaystyle \begin{aligned}x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}&\geq x\left(1+\frac{y^{x-1}}{x^y}\right)\\&\geq x\left(1+\frac{x^{y-1}}{x^y}\right)\\&=x+1>1\end{aligned}

and the proof is completed Q.E.D.

# Inequality 54 (George Basdekis)

Problem:

Let $a,b$ and $c$ be real numbers. Prove the Inequality

$\displaystyle\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq\frac{9}{2(a^2+b^2+c^2)}.$

Solution:

Without loss of generality, let us assume that $a>b>c$. Then according to the AM-GM Inequality we have that

\displaystyle\begin{aligned}\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}\geq\frac{2}{(a-b)(b-c)}&\geq\frac{2\cdot 4}{[(a-b)+(b-c)]^2}\\&=\frac{8}{(a-c)^2}.\end{aligned}

Therefore, it holds that

$\displaystyle\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq\frac{9}{(c-a)^2}.$

It remains now to prove that

$\displaystyle \frac{9}{(c-a)^2}\geq\frac{9}{2(a^2+b^2+c^2)},$

which reduces to the obvious, after expansions, Inequality $(a+c)^2+2b^2\geq 0.$ Equality holds if and only if $(a,b,c)\sim (-b,b,0)$ or any cyclic permutations of the previous equality Q.E.D.

# The blog is active again

Old loves cannot and definitely must not be forgotten so I decided to make my blog here active again. I will try to post Inequalities from around the world. Happy sharing Inequalities.

George.

# Inequality 53 (Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c>0$ such that $\displaystyle a+b+c=3$ and $\displaystyle a^2+b^2+c^2=4$. Find the minimum and the maximum value of the expression

$\displaystyle Q=\frac{a}{b}$.

Solution:

From the Cauchy – Schwarz Inequality we have that

$\displaystyle (a^2+b^2+c^2)\cdot\left[\frac{(a+b)^2}{a^2+b^2}+1\right]\geq (a+b+c)^2$,

or due to the hypothesis

$\displaystyle \frac{(a+b)^2}{a^2+b^2}\geq \frac{5}{4}$.

From this last Inequality, we acquire

$\displaystyle a^2+b^2\leq 8ab$,

Let us now divide with $\displaystyle ab$. Then we get that

$\displaystyle \frac{a}{b}+\frac{b}{a}\leq 8$.

Solving this Inequality, gives us

$\displaystyle 4-\sqrt{15}\leq \frac{a}{b}\leq 4+\sqrt{15}$.

Thus, $\displaystyle Q_{\min}=4-\sqrt{15}$ and $\displaystyle Q_{\max}=4+\sqrt{15}$, Q.E.D.

# Inequality 52 (Unknown Author)

Problem:

Let $\displaystyle x_1,x_2,...,x_n$ be positive real numbers with sum equal to $\displaystyle 1$. Find the minimum value of the expression

$\displaystyle A=\max\left\{\frac{x_1}{1+x_1},\frac{x_2}{1+x_1+x_2},...,\frac{x_n}{1+x_1+x_2+...+x_n}\right\}$.

Solution:

From the definition of $\displaystyle A$, we have that

$\displaystyle A\geq \frac{x_1}{1+x_1}, A\geq \frac{x_2}{1+x_1+x_2},...,A\geq \frac{x_n}{1+x_1+x_2+...+x_n}$.

Solving the 1st Inequality we get that

$\displaystyle x_1\leq \frac{A}{1-A}$.

Solving now the 2nd Inequality we have that

$\displaystyle A\geq \frac{x_2}{1+x_1+x_2}\geq \frac{x_2}{1+\frac{A}{1-A}+x_2}=\frac{x_2}{\frac{1}{1-A}+x_2}$.

The last relation reduces to the

$\displaystyle x_2\leq \frac{A}{(1-A)^2}$.

So, we see now that $\displaystyle x_k\leq \frac{A}{(1-A)^k}$ for $\displaystyle k=1,2,...,n$.

Summing up these $\displaystyle n$ Inequalities we acquire

$\displaystyle x_1+x_2+...+x_n\leq \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}$

or, due to the hypothesis we now have that

$\displaystyle \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}\geq 1$.

If we solve this last Inequality it gives us the result $\displaystyle A\geq 1-\frac{1}{\sqrt[n]{2}}$, which is also and the value we are searching, Q.E.D.

# Inequality 51 (George Basdekis)

Problem:

If $\displaystyle a,b,c>0$ then prove that

$\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \frac{3}{\sqrt{2}}\cdot\sqrt{\frac{a^2+b^2+c^2}{a+b+c}}$.

Solution:

From the Cauchy – Schwarz Inequality we have that

$\displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \left[\sum a(b+c)\right]\cdot\sum\frac{a}{(a+b)(b+c)}$.

Expanding the RHS, we get that

\displaystyle \begin{aligned}\left(\sum\frac{a}{\sqrt{a+b}}\right)^2& \leq 2(ab+bc+ca)\cdot \frac{a^2+b^2+c^2+ab+bc+ca}{(a+b)(b+c)(c+a)}\\&\leq 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\end{aligned},

from the obvious $\displaystyle ab+bc+ca\leq a^2+b^2+c^2$.

Moreover, notice that the following beautiful Inequality holds, that is

$\displaystyle 9(a+b)(b+c)(c+a)\geq 8(a+b+c)(ab+bc+ca)$.

Thus, we have that

$\displaystyle 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}$.

We have proved our Inequality, because $\displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}$, or

$\displaystyle \sum\frac{a}{\sqrt{a+b}}\leq 3\sqrt{\frac{a^2+b^2+c^2}{2(a+b+c)}}$, Q.E.D.

# Inequality 50 (Vasile Cirtoaje)

Problem:

For all $\displaystyle a,b,c>0$ prove that

$\displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}+\frac{3b^2-2bc-c^2}{b^2+c^2}+\frac{3c^2-2ca-a^2}{c^2+a^2}\geq 0$.

Solution:

It holds that $\displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}=\frac{2(a^2-b^2)+(a-b)^2}{a^2+b^2}$. Thus, we only need to prove that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq -2\sum\frac{a^2-b^2}{a^2+b^2}$.

Observe that

$\displaystyle \sum\frac{a^2-b^2}{a^2+b^2}=-\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

From the above result we have to prove now that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

Using the AM – GM Inequality we get that

$\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$.

So, it is enough to prove that

$\displaystyle 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$.

The last Inequality can be reduced to the $\displaystyle 27(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2\geq 8(a-b)(b-c)(c-a)(a+b)^3(b+c)^3(c+a)^3$.

So, it is enough to check the Inequality

$\displaystyle 3(a^2+b^2)^2\geq 2|a-b|(a+b)^3$.

This one can be prove by the following way. Set $\displaystyle a^2+b^2=x$ and $\displaystyle a^2-b^2=y$, with $\displaystyle x\geq y$ and remake the Inequality to the form

$\displaystyle 3\frac{(a^2+b^2)^2}{(a+b)^2}\geq 2|a^2-b^2|$.

Then, the Inequality substitutes to

$\displaystyle 3\frac{x^2}{x+\sqrt{x^2-y^2}}\geq 2y$.

But this one holds because we have that

$\displaystyle 2xy++2y\sqrt{x^2-y^2}\leq 2xy+(x^2-y^2)+y^2=x^2+2xy\leq 3x^2$,

due to the AM – GM Inequality and the hypothesis $\displaystyle x \geq y$, Q.E.D.

P.S The following nice Inequality also holds:

$\displaystyle \sum \frac{3a^2-2ab-b^2}{3a^2+2ab+3b^2}\geq 0$.

This Inequality belongs to Thomas Mildorf and the proof for this Inequality is the same as the above.

# Inequality 49 (Unknown Author)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle a+b+c=3$, and also let $\displaystyle n\geq 12$ be a natural number. Prove that

$\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{3}{\sqrt[n-4]{n^n}}$.

Solution:

Observe that $\displaystyle \frac{1}{3a+bc+\frac{n-4}{a}}=\frac{a}{3a^2+bc+(n-4)}=\frac{a}{a(a+b)(a+c)+(n-4)}$. Thus, from the AM – GM Inequality we get that

$\displaystyle 4\cdot\frac{a(a+b)(a+c)}{4}+(n-4)\geq n\sqrt[n]{\frac{a^4(a+b)^4(a+c)^4}{4^4}}$.

From the above Inequality, we are now capable of building our problem. Specifically, we have that

$\displaystyle \frac{a}{a(a+b)(a+c)+(n-4)}\leq \frac{1}{n}\sqrt[n]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}$

or

$\displaystyle \left(\frac{a}{a(a+b)(a+c)+(n-4)}\right)^{\frac{n}{n-4}} \leq \frac{1}{\sqrt[n-4]{n^n}}\sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}$.

We must see now that

$\displaystyle \sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}=\sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}$.

From the above result we can apply once again the AM – GM Inequality and acquire

$\displaystyle \sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}\leq \frac{4\frac{2a}{a+b}+4\frac{2a}{a+c}+(n-12)a}{n-4}$.

Summing up the $\displaystyle 3$ Inequalities together we get that

$\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}$.

Moreover, we have that

$\displaystyle \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}=\frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{3\cdot 4\cdot 2+3(n-12)}{n-4}$,

which reduces to the $\displaystyle \frac{3}{\sqrt[n-4]{n^n}}$, Q.E.D.