Putting a= b=c won't work

Algebra Level 5

Let a,b,c be positive real numbers so that $a^{4} + b^{4} +c^{4} =4$

Then let the maximum value of the expression $a^{2} b^{2} \sqrt{7} +c^{2} b^{2} 3$ be m and the maximum value of the expression $a \times c^{2} \times b$ be $\sqrt{n}$

Give your answer as m×n

The answer is 16.

1 solution

Manuel Kahayon
Feb 22, 2016

By AM-GM

$(\frac{a^4+\frac{7b^4}{7+9}}{2})+(\frac{\frac{9b^4}{7+9}+c^4}{2}) \geq \sqrt{\frac{7a^4b^4}{7+9}}+\sqrt{\frac{9b^4c^4}{7+9}}$

Which gives us $8 \geq \sqrt{7}a^2b^2+3b^2c^2$ when simplified.

Also, $\frac{a^4+b^4+\frac{c^4}{2}+\frac{c^4}{2}}{4} \geq \sqrt[4]{\frac{a^4b^4c^8}{4}}$

Which gives us $\sqrt{2} \geq abc^2$ when simplified.

So, our answer is $8 \cdot 2 = \boxed {16}$ .

Putting a= b=c won't work

The answer is 16.

1 solution

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