Almost symmetry!

Algebra Level 4

$\large \frac { \frac { { a }^{ 4 } }{ bc } +\frac { { b }^{ 4 } }{ ac } +\frac { { c }^{ 4 } }{ ab } }{ 8 } =\frac { { a }^{ 5 }+{ b }^{ 5 }+{ { c }^{ 5 } } }{ 2{ c }^{ 4 }+{ b }^{ 4 }+2{ a }^{ 3 }+4a }$

If $a,b$ and $c$ are positive real numbers that satisfy the equation above, find the value of $a^2$ .

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1 solution

Adam Pet
Nov 11, 2015

By AM-GM , we have that

$2c^4 + b^4 + 2a^3 + 4a \geq 8abc$

Multiplying the left side by $8abc$ and the right side by $2c^4 + b^4 + 2a^3 + 4a$ yields $a^5 + b^5 + c^5 = a^5 + b^5 + c^5$ . Therefore,

$\frac{\frac{a^4}{bc} + \frac{b^4}{ac} + \frac{c^4}{ab}}{8} \geq \frac{a^5 + b^5 + c^5}{2c^4 + b^4 + 2a^3 + 4a}$

with equality holding when $2c^4 = b^4 = 2a^3 = 4a$ , hence $a^2 = 2$ .

In the last line there is a little mistake: 2c^4=b^4=2a^3=4a.

Osvaldo Guimaraes - 5 years, 6 months ago

Edited, thanks for your observation!

Adam Pet - 5 years, 6 months ago

Almost symmetry!

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