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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.

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