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# Inequality 17(Darij Grinberg)

Problem:

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

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

Solution:

Let us multiply each fraction cyclic by $\displaystyle a^2(b+c), b^2(c+a), c^2(a+b)$ respectively. Then we acquire that $\displaystyle \sum_{cyc}\frac{a^4(b+c)^2}{a^2(b+c)(b^2+c^2)}\geq \sum_{cyc}a$.

Now from Cauchy-Schwartz inequality we get

$\displaystyle \sum_{cyc}\frac{a^4(b+c)^2}{a^2(b+c)(b^2+c^2)}\geq \frac{\left(\sum_{cyc}a^2(b+c)\right)^2}{\sum_{cyc}a^2(b+c)(b^2+c^2)}$.

So, we only need to prove that

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

or

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

Let us denote by $\displaystyle p,q$ the $\displaystyle a+b+c,ab+bc+ca$, that is $\displaystyle p=a+b+c, q=ab+bc+ca$.

We have that: $\displaystyle \left(\sum_{cyc}a^2(b+c)\right)^2=\left(\sum_{cyc}a\sum_{cyc}ab-3abc\right)^2=(pq-3abc)^2=p^2q^2+9a^2b^2c^2-6abcpq$$\displaystyle (1)$.

On the other hand:

$\displaystyle \sum_{cyc}a^2(b+c)(b^2+c^2)=\sum_{cyc}(b+c)(a^2b^2+b^2c^2+c^2a^2-b^2c^2)=\sum_{cyc}(b+c)\sum_{cyc}a^2b^2-\sum_{cyc}(b+c)b^2c^2$.

But $\displaystyle b+c=p-a$. So,

$\displaystyle \sum_{cyc}(b+c)\sum_{cyc}a^2b^2-\sum_{cyc}(b+c)b^2c^2=2p\sum_{cyc}a^2b^2-\sum_{cyc}(p-a)b^2c^2$.

So, we finally have that

$\displaystyle p\sum_{cyc}a^2b^2-abcq=p(q^2-2abcp)-abcq=pq^2-2abcp^2+abcq$.

Multiplying by $\displaystyle p$ we aqcuire $\displaystyle p^2q^2-2abcp^3+abcpq$, $\displaystyle (2)$.

From $\displaystyle (1),(2)$ we deduce that

$\displaystyle 2p^3+9abc\geq 7pq$.

But from Schur’s inequality we know that

$\displaystyle p^3+9abc\geq 4pq$, and $\displaystyle p^3\geq p^2\geq 3q$.

Adding up these $\displaystyle 2$ inequalities we kill our inequality, Q.E.D.