# Inequality 47(Christos Patilas)

Problem:

If $\displaystyle x_i$ for $\displaystyle i=1,2,...,n$ are positive real numbers then prove that

$\displaystyle \sum^{n}_{i=1}\left(5\sqrt[5]{x^{3}_{i}}-3\sqrt[3]{\left(\frac{3x_i+2}{5}\right)^5}\right)\leq 2n$

Solution(An Idea by Vo Quoc Ba Can):

We only need to prove that

$\displaystyle 5\sqrt[5]{a^3}-3\sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\leq 2$ for all $\displaystyle a>0$.

So, using the AM-GM Inequality we have that

$\displaystyle a+a+a+1+1\geq 5\sqrt[5]{a^3}$.

It follows that

$\displaystyle \sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\geq \sqrt[3]{\left(\sqrt[5]{a^3}\right)^5}=a$.

Therefore it suffices to prove that

$\displaystyle 5\sqrt[5]{a^3}-2\leq 3a$,

which is obviously true from the AM-GM Inequality, Q.E.D.

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# Inequality 46(George Basdekis)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that $\displaystyle \left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)-8abc\geq a(b-c)^2+b(c-a)^2+c(a-b)^2$.

# Inequality 45(Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle ab+bc+ca=1$. Prove that $\displaystyle \frac{1}{\sqrt{1+(2a-b)^2}}+\frac{1}{\sqrt{1+(2b-c)^2}}+\frac{1}{\sqrt{1+(2c-a)^2}}\leq\frac{3\sqrt{3}}{2}$.

# Inequality 44(Unknown Author)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that

$\displaystyle 1+\frac{8abc}{(a+b)(b+c)(c+a)}\geq \frac{2(ab+bc+ca)}{a^2+b^2+c^2}$.

Solution (An idea by Silouanos Brazitikos):

From the above inequality is it enough to show that

$\displaystyle \frac{8abc}{(a+b)(b+c)(c+a)}\geq \frac{2(ab+bc+ca)-a^2+b^2+c^2}{a^2+b^2+c^2}$.

But from Schur’s Inequality we know that

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

So it is enough to check that

$\displaystyle 8(a^2+b^2+c^2)(a+b+c)\geq 9(a+b)(b+c)(c+a)$.

From Cauchy-Schwarz Inequality we know that

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

Therefore we only need to prove that

$\displaystyle \left(\frac{a+b+b+c+c+a}{3}\right)^{3}\geq (a+b)(b+c)(c+a)$,

which is obviously true from AM-GM Inequality, Q.E.D.

# Inequality 43(Vo Quoc Ba Can & George Basdekis)

Problem:

If $\displaystyle a,b,c$ are positive real numbers satisfying the equality $\displaystyle abc=1$, then prove that

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

Solution:

Without loss of generality, assume that $\displaystyle a\geq b\geq c$.

Since $\displaystyle a^{\frac{2}{3}}\geq b^{\frac{2}{3}}\geq c^{\frac{2}{3}}$

and

$\displaystyle \frac{1}{\sqrt{b+c}}\geq\frac{1}{\sqrt{c+a}}\geq \frac{1}{\sqrt{a+b}}$,

using Chebyshev’s Inequality we get:

$\displaystyle \frac{a^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{b^{\frac{2}{3}}}{\sqrt{c+a}}+\frac{c^{\frac{2}{3}}}{\sqrt{a+b}}\geq \frac{1}{3}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}\right)\cdot\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}+\frac{1}{\sqrt{a+b}}\right)$.

Moreover, From the AM-GM Inequality we have that

$\displaystyle \frac{1}{3}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}\right)\cdot\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}+\frac{1}{\sqrt{a+b}}\right)\geq \frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}}{\sqrt[6]{(a+b)(b+c)(c+a)}}$.

Therefore, it suffices to prove that

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

Set $\displaystyle a=x^3,b=y^3,c=z^3$.

The above inequality can be written now as

$\displaystyle 2(x^2+y^2+z^2)^2\geq 9\sqrt[3]{(x^3+y^3)(y^3+z^3)(z^3+x^3)}$, or $\displaystyle 2(x^2+y^2+z^2)^2\geq 9\sqrt[3]{xyz(x^3+y^3)(y^3+z^3)(z^3+x^3)}$.

From the AM-GM Inequality once again, we acquire that

$\displaystyle \sqrt[3]{xyz(x^3+y^3)(y^3+z^3)(z^3+x^3)}\leq \frac{x(y^3+z^3)+y(z^3+x^3)+z(x^3+y^3)}{3}$.

And thus, it is enough to check that

$\displaystyle 2(x^2+y^2+z^2)^{2}\geq 3\left[x(y^3+z^3)+y(z^3+x^3)+z(x^3+y^3)\right]$,

which is equivalent to the obvious inequality

$\displaystyle (x^2-xy+y^2)(x-y)^2+(y^2-yz+z^2)((y-z)^2+(z^2-zx+x^2)(z-x)^2\geq 0$, Q.E.D.

# Inequality 42(George Basdekis)

Problem:

If $\displaystyle a,b,c$ are positive real numbers such that $\displaystyle abc=1$, then prove that

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

Solution:

From the Cauchy-Schwarz inequality we deduce that

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

Removing the square we get that

$\displaystyle \sqrt{a^3+b^3}\cdot\sqrt{a+b}\geq (a^2+b^2)$. Let us now divide by $\displaystyle a^2+b^2$.

Then we have $\displaystyle \frac{\sqrt{a^3+b^3}}{a^2+b^2}\geq \frac{1}{\sqrt{a+b}}$.

Moreover, multiply by $\displaystyle c$. We, thus, acquire

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

So,we have proved that

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

We will now apply Holder’s Inequality, that is

$\displaystyle \left(\frac{c}{\sqrt{a+b}}+\frac{b}{\sqrt{c+a}}+\frac{a}{\sqrt{b+c}}\right)^2\cdot\left[c(a+b)+b(c+a)+a(b+c)\right]\geq \left(a+b+c\right)^3$, or $\displaystyle \left(LHS\right)^2\geq \frac{(a+b+c)^3}{2(ab+bc+ca)}$.

Rewrite the sum $\displaystyle (a+b+c)^3$ as $\displaystyle (a+b+c)^2\cdot(a+b+c)$.

Then we get that:

\displaystyle \begin{aligned}\frac{(a+b+c)^3}{2(ab+bc+ca)}=\frac{(a+b+c)^2\cdot(a+b+c)}{2(ab+bc+ca)}&\geq \frac{3(ab+bc+ca)(a+b+c)}{2(ab+bc+ca)}\\&=\frac{3(a+b+c)}{2}\end{aligned}.

And finally, from the AM-GM inequality we have

$\displaystyle \frac{ 3(a+b+c)}{2}\geq \frac{3\cdot 3\sqrt[3]{abc}}{2}=\frac{9}{2}$

So, we have proved that

$\displaystyle \left(LHS\right)^2\geq \frac{9}{2}\Longrightarrow LHS\geq \frac{3}{\sqrt{2}}\Longrightarrow\frac{c\sqrt{a^3+b^3}}{a^2+b^2}+\frac{a\sqrt{b^3+c^3}}{b^2+c^2}+\frac{b\sqrt{c^3+a^3}}{c^2+a^2}\geq \frac{3}{\sqrt{2}}$, Q.E.D.

# Inequality 41(Christos Patilas)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle a+b+c=3$. Prove that

$\displaystyle b\cdot\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+c\cdot\frac{(a+2b+c)^2}{2b^2+(c+a)^2}+a\cdot\frac{(a+b+2c)^2}{2c^2+(a+b)^2}\leq 8$.

Solution:

From the hypothesis, the inequality is of the form $\displaystyle b\cdot\frac{(a+3)^2}{2a^2+(3-a)^2}+c\cdot\frac{(b+3)^2}{2b^2+(3-b)^2}+a\cdot\frac{(c+3)^2}{2c^2+(3-c)^2}\leq 8$. If we expand the nominators and the denominators, then we get that

$\displaystyle b\cdot\frac{a^2+6a+9}{3a^2-6a+9}+c\cdot\frac{b^2+6b+9}{3b^2-6b+9}+a\cdot\frac{c^2+6c+9}{3c^2-6c+9}\leq 8$.

Now let us use once again the Cauchy-Reverse technique.

$\displaystyle \frac{a^2+6a+9}{3a^2-6a+9}=\frac{1}{3}\cdot\frac{3a^2+18a+27}{3a^2-6a+9}=\frac{1}{3}\cdot\frac{(3a^2-6a+9)+(24a+18)}{3a^2-6a+9}$.

Thus, we have

$\displaystyle \frac{1}{3}\cdot\left(1+\frac{24a+18}{3a^2-6a+9}\right)$.

Moreover,

$\displaystyle 3a^2-6a+9=3(a-1)^2+6\geq 6\Longrightarrow \frac{1}{3a^2-6a+9}\leq\frac{1}{6}$.

So,

$\displaystyle \frac{1}{3}\cdot\left(1+\frac{24a+18}{3a^2-6a+9}\right)\leq \frac{1}{3}\cdot\left(1+\frac{24a+18}{6}\right)$.

Multiplying by $\displaystyle b$ we acquire

$\displaystyle b\cdot\frac{1}{3}\cdot\left(1+\frac{24a+18}{3a^2-6a+9}\right)\leq b\cdot\frac{1}{3}\cdot\left(1+\frac{24a+18}{6}\right)=b\cdot\left(\frac{1}{3}+\frac{8a+6}{6}\right)$.

Now, if we sum up the $\displaystyle 3$ inequalities it remains to prove that

$\displaystyle LHS\leq \sum_{cyc}b\cdot\left(\frac{1}{3}+\frac{8a+6}{6}\right)=\frac{1}{3}\sum_{cyc}b+\frac{8}{6}\sum_{cyc}ab+\sum_{cyc}b\leq 8$,

which reduces to the obvious inequality $\displaystyle \sum_{cyc}ab\leq 3$, Q.E.D.