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# 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.