work 023 – mathlinks user socrates

Problem

Let  a,b,c > 0. Prove that:
\frac {\sqrt {a^3 + b^3}}{a^2 + b^2} + \frac {\sqrt {b^3 + c^3}}{b^2 + c^2} + \frac {\sqrt {c^3 + a^3}}{c^2 + a^2}

\geq \frac {6(ab + bc + ca)}{(a + b + c)\sqrt {(a + b)(b + c)(c + a)}}

Proof

By C-S ,
\frac {\sqrt {a^3 + b^3}}{a^2 + b^2} \ge \frac {1}{\sqrt {a + b}}
Thus it is sufficient to prove that (after removing \sqrt {(a + b)(a + c)(b + c)} in the denominators)-
\sum \sqrt {(b + c)(b + a)} \ge \frac {6(ab + bc + ca)}{a + b + c}
Let ,
x^2 = a + b ,y^2 = b + c , z^2 = c + a
The inequality is transformed into –

\sum xy \ge 3\cdot \frac{2(\sum x^2y^2) - \sum x^4}{x^2+y^2+z^2}

(x^2+y^2+z^2)(xy+yz+zx) + 3(x^4+y^4+z^4)

\ge 6(x^2y^2 + y^2z^2 + z^2x^2)

\Leftrightarrow \sum x^3(y + z) + \sum x^4 + xyz(x + y + z) +

2 \sum (x^4 - x^2y^2)

\ge 4(\sum x^2y^2)

By AM-GM ,

\sum x^4 \ge \sum x^2y^2

Thus it is sufficient to prove that –

\sum x^3(x + y + z) + xyz(x + y + z) \ge 4\sum x^2y^2

\Leftrightarrow (x^3 + y^3 + z^3 + xyz)(x + y + z) \ge 4(x^2y^2 + y^2z^2 + z^2x^2)

By Schur’s Inequality we know that –

\sum x^3 + 3xyz \ge \sum x(y^2 + z^2)

Thus it is left to prove that –

(\sum x(y^2 + z^2) - 2xyz)(x + y + z) \ge 4(x^2y^2 + y^2z^2 + z^2x^2)

This is equivalent to –

\sum xy(x^2 + y^2) \ge 2 \sum x^2y^2

\Leftrightarrow \sum xy(x - y)^2 \ge 0
Hence Proved with equality when all the variables are equal  🙂

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