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# Inequality 2(Vasile Cirtoaje)

Problem:

Let $\displaystyle a,b,c$ be non-negative numbers, no two of them are zero. Prove that

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

1st solution:

Let $\displaystyle A=a^2+ab+b^2,B=b^2+bc+c^2,C=c^2+ca+a^2$. We have

\displaystyle \begin{aligned}\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\left(\frac{a^2}{A}+\frac{b^2}{B}+\frac{c^2}{C}-1\right)=\sum_{cyc}\frac{a^2}{A^2}+\sum_{cyc}\frac{b^2+c^2}{BC}-\sum_{cyc}\frac{1}{A}&=\frac{1}{2}\sum_{cyc}\left(\frac{b}{B}-\frac{c}{C}\right)^{2}\\&\geq 0\end{aligned}

from which the desired inequality follows. Equality occurs if and only if $\displaystyle a=b=c$.

2nd solution:

Divide each fraction with $\displaystyle a^2,b^2,c^2$ respectively. Then we get that

$\displaystyle \sum_{cyc}\frac{1}{1+\frac{b}{a}+\left(\frac{b}{a}\right)^{2}}\geq 1$.

Let us denote $\displaystyle \frac{b}{a}=x,\frac{c}{b}=y,\frac{a}{c}=z$. Then the inequality transforms to

$\displaystyle \sum_{cyc}\frac{1}{x^2+x+1}\geq 1$.

Let us now use the transformation $\displaystyle x=\frac{uv}{w^2},y=\frac{vw}{u^2},z=\frac{wu}{v^2}$ which makes the inequality to

$\displaystyle \sum_{cyc}\frac{u^{4}}{u^4+v^2w^2+u^2vw}\geq 1$.

From Cauchy’s inequality now, we get that

$\displaystyle \sum_{cyc}\frac{u^{4}}{u^4+v^2w^2+u^2vw}\geq \frac{\left(\sum_{cyc}u^{2}\right)^{2}}{\sum_{cyc}(u^4+u^2v^2)+uvw\sum_{cyc}u}$.

So, we only need to prove that

$\displaystyle \frac{\left(\sum_{cyc}u^{2}\right)^{2}}{\sum_{cyc}(u^4+u^2v^2)+uvw\sum_{cyc}u}\geq 1\Longleftrightarrow \sum_{cyc}u^4\geq uvw\sum_{cyc}u$,

which is obviously true. Equality holds only for $\displaystyle a=b=c$, Q.E.D.