# 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.

# Inequality 48 (George Basdekis)

Problem:

If $\displaystyle a,b,c,d,e,f\geq 0$ such that $\displaystyle a+b+c+d+e+f=6$ and $\displaystyle a^2+b^2+c^2+d^2+e^2+f^2=\frac{36}{5}$, the prove that

$\displaystyle a^3+b^3+c^3+d^3+e^3+f^3\leq \frac{264}{25}$.

Solution:

Using the Cauchy – Schwarz Inequality we get that

$\displaystyle \frac{36}{5}=a^2+(b^2+...+f^2)\geq a^2+\frac{(b+...+f)^2}{5}=a^2+\frac{(6-a)^2}{5}=\frac{36}{5}+\frac{6a^2-12a}{5}$.

From the above Inequality we can see that $\displaystyle a\leq 2$. Similarly we acquire $\displaystyle b,...,f\leq 2$.

Observe now that

$\displaystyle \sum_{cyc} a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a)$.

Using once again the Cauchy – Schwarz Inequality we get that

$\displaystyle \sum_{cyc} a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}$.

Using the hypothesis, we can calculate the last fraction. Thus we have that

$\displaystyle \sum_{cyc}a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}=\frac{96}{25}$.

Returning to previous Inequality, we now can see that

$\displaystyle \sum_{cyc}a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a)\leq 2\cdot \frac{36}{5}-\frac{96}{25}=\frac{264}{25}$, Q.E.D.