# Inequality 15(Kvant)

Problem:

Prove that for any real numbers $\displaystyle a_1,a_2,...,a_n$ the following inequality holds:

$\displaystyle \left(\sum^{n}_{i=1}a_{i}\right)^{2}\leq \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}$.

Solution:

Observe that

$\displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}=\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt$.

But $\displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}$ can be considered as a constant. So,

$\displaystyle \sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt=\int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt$.

Notice now that

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt=\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt$.

So, the inequality reduces to

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\sum^{n}_{i=1}a_{i}\right)^{2}$.

Now, using Cauchy-Schwartz inequality for integrals, we get that

$\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}$.

But, we must now observe that

$\displaystyle \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}=\left(\sum^{n}_{i=1}a_{i}\right)^{2}$,

which comes to the conclusion, Q.E.D.

# Inequality 14(Unknown Author)

Problem:

Let $\displaystyle a,b,c>0$ with $\displaystyle a+b+c=1$. Show that

$\displaystyle \frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq 2$.

Solution:

Subtract each side with $\displaystyle -1$. Then, from hypothesis we have that

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

Or

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

From the last relation we have that

$\displaystyle \sum_{cyc}\frac{a(a-c)}{b+c}+\sum_{cyc}a\geq 1\Longrightarrow \sum_{cyc}\frac{a(a-c)}{b+c}\geq 0$.

Assume without loss of generality that $\displaystyle 0.

Then we have that

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

Doing some manipulations on both sides we acquire that

$\displaystyle b^4+b^3c-a^2b^2-a^2bc+c^4+c^3a-b^2c^2-ab^2c\geq a^2c^2+abc^2-a^4-a^3b$.

Or,

$\displaystyle \sum_{cyc}a^4+\sum_{cyc}a^3b\geq \sum_{cyc}a^2b^2+abc\sum_{cyc}a$,

which holds.

Indeed, from the AM-GM inequality we have that

$\displaystyle \sum_{cyc}a^4\geq \sum_{cyc}a^2b^2$

and from Cauchy-Schwartz inequality

$\displaystyle \sum_{cyc}\frac{a^2}{c}\geq \sum_{cyc}a\Longrightarrow \sum_{cyc}a^3b\geq abc\sum_{cyc}a$.

Adding up these $\displaystyle 2$ inequalities we get the desired result, Q.E.D.

# Inequality 13(Silouanos Brazitikos & Christos Patilas)

The problem is locked until April 2, 2010. Until then be patient my friends.

Regards, George.

# Inequality 12(Vasile Cirtoaje)

Problem:

If $\displaystyle a,b,c$ are non-negative numbers prove that

$\displaystyle (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\geq (ab+bc+ca)^3$.

Solution:

Lemma: $\displaystyle 4(a^2+ab+b^2)\geq 3(a+b)^2$.

Back to the inequality now, multiply both sides by $\displaystyle 64$. Then we have that $\displaystyle 4^3\prod_{cyc}(a^2+ab+b^2)\geq 4^3(ab+bc+ca)^3$.

But from the lemma we reduce the current inequality to

$\displaystyle 27\prod_{cyc}(a+b)^2\geq 64(ab+bc+ca)^3$.

It also holds $\displaystyle (a+b+c)^2\geq 3(ab+bc+ca)$. Multiplying the last inequality with $\displaystyle \frac{64}{3}(ab+bc+ca)^2$ we get that $\displaystyle \frac{64}{3}(ab+bc+ca)^2(a+b+c)^2\geq 64(ab+bc+ca)^3$.

So, it suffices to prove that

$\displaystyle 27\prod_{cyc}(a+b)^2\geq \frac{64}{3}(a+b+c)^2(ab+bc+ca)^2$

or

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

which reduces to the obvious inequality

$\displaystyle a(b-c)^2+b(c-a)^2+c(a-b)^2\geq 0$.

Equality occurs for $\displaystyle (a,b,c)=(1,1,1)$ and also for $\displaystyle (a,b,c)=(1,0,0)$ or any cyclic permutation, Q.E.D.

# Inequality 11(2009 Mediterranean Mathematical Olympiad)

Problem:

If $\displaystyle x,y,z$ are positive real numbers prove that

$\displaystyle \frac{xy}{xy+x^2+y^2}+\frac{yz}{yz+y^2+z^2}+\frac{zx}{zx+z^2+x^2}\leq \frac{x}{2x+z}+\frac{y}{2y+x}+\frac{z}{2z+y}$.

Solution (An idea by Vo Quoc Ba Can):

Using the Cauchy-Schwarz Inequality we have that

$\displaystyle \frac{x}{2x+z}+\frac{y}{2y+x}+\frac{z}{2z+y}=\frac{x^2}{2x^2+xz}+\frac{y^2}{2y^2+yx}+\frac{z^2}{2z^2+zy}\geq \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$.

Thus it suffices to show that

$\displaystyle \sum_{cyc}\frac{xy}{x^2+xy+y^2}\leq \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$,

which is equivalent to

$\displaystyle \sum_{cyc}\left(\frac{1}{3}-\frac{xy}{x^2+xy+y^2}\right)\geq 1-\frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$,

or

$\displaystyle \sum_{cyc}\frac{(x-y)^2}{3(x^2+xy+y^2)}\geq \frac{x^2+y^2+z^2-xy-yz-zx}{2(x^2+y^2+z^2)+xy+yz+zx}$.

Since $\displaystyle x^2+y^2+z^2-xy-yz-zx=\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}$ the above inequality can be rewritten as

$\displaystyle \sum_{cyc}(x-y)^2\left[\frac{1}{3(x^2+xy+y^2)}-\frac{1}{2(x^2+y^2+z^2)+xy+yz+zx}\right]\geq 0$,

which is obviously true, Q.E.D.

# Inequality 10(Christos Patilas)

Problem:

If $\displaystyle a,b,c\in \mathbb{R}$, prove that

$\displaystyle \sqrt[4]{a^4+b^4+c^4+1}\geq \sqrt[5]{a^5+b^5+c^5+1}$.

Solution:

We have that

$\displaystyle \sqrt[5]{\sum_{cyc}a^5+1}\leq \sqrt[5]{\sum_{cyc}\left|a\right|^5+1}$,

so we only need to prove the inequality for $\displaystyle a,b,c\geq 0$.

Let us write the given inequality into the form

$\displaystyle \left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}\geq \sum_{cyc}a^5+1$ and divide with $\displaystyle \left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}$.

Then we acquire

$\displaystyle \sum_{cyc}\frac{a^5}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq 1$

or

$\displaystyle \sum_{cyc}\left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq 1$.

But for all non-negative numbers holds that

$\displaystyle \left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}\leq \frac{a^4}{\sum_{cyc}a^4+1}$.

So from the above inequality we get that

$\displaystyle \sum_{cyc}\left(\frac{a^4}{\sum_{cyc}a^4+1}\right)^{\frac{5}{4}}+\frac{1}{\left(\sum_{cyc}a^4+1\right)^{\frac{5}{4}}}\leq \frac{\sum_{cyc}a^4+1}{\sum_{cyc}a^4+1}=1$, Q.E.D.

# Inequality 9(Christos Patilas)

Problem:

If $\displaystyle a_{1},a_{2},a_{3}$ are the positive real roots of the equation $\displaystyle 4x^3-kx^2+mx-9=0$ prove that

$\displaystyle k\geq 4\sqrt[3]{\sum_{cyc}\left(a_{1}\sqrt{a_{2}+a_{3}}\right)+3\prod_{cyc}(a_{1}+a_{2})}$.

Solution:

Let us divide both sides by $\displaystyle 4$ and then cube them.

We acquire

$\displaystyle \left(\frac{k}{4}\right)^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$.

But from Viete’s relations we have that

$\displaystyle \frac{k}{4}=a_{1}+a_{2}+a_{3}$ and $\displaystyle a_{1}a_{2}a_{3}=\frac{9}{4}$.

So our inequality transforms into

$\displaystyle (a_{1}+a_{2}+a_{3})^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$,

or

$\displaystyle \sum_{cyc} a^{3}_{1}+3\prod_{cyc}(a_{1}+a_{2}) \geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2})$.

So, it suffices to prove that

$\displaystyle \sum_{cyc} a^{3}_{1}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}$.

But the last inequality holds because

$\displaystyle a^{3}_{1}+a^{3}_{2}+a^{3}_{3}\geq a_{1}a_{2}(a_{1}+a_{2})+a^{3}_{3}\geq 2\sqrt{a_{1}a_{2}a_{3}\cdot a^{2}_{3}(a_{1}+a_{2})}=3a_{3}\sqrt{a_{1}+a_{2}}$.

Adding up the $\displaystyle 3$ cyclic relations we come to the desired inequality, Q.E.D.