Where did the $2$ come from?

Algebra Level 5

$\large K\sqrt{N}\le (a+2b^{2}+2c^{2})$

The above inequality holds for all positive real numbers $a,b,c$ . Here $K$ is a constant and $N=abc$ .

When equality occurs,let the value of $\dfrac{a}{b^{2}}=x$

Find $K^{x}$ .

The answer is 256.

1 solution

Raven Herd
Mar 7, 2016

a+ $2b^2$ + $2c^2$ = $\frac{a}{2}$ + $\frac{a}{2}$ + $2b^2$ + $2c^2 \ - By Arithmetic Mean Inequality , \(\frac{a}{2}$ + $\frac{a}{2}$ + $2b^2$ + $2c^2 \ >= 4(\(\frac{a}{2}$ $\frac{a}{2}$ $2b^2$ * $2c^2$ ^1/4
>=4 $(abc)^1/2$
Equality occurs when $\frac{a}{2}$ = $\frac{a}{2}$ = $2b^2$ = (2c^2 \
x = 4
k=4
k^x = 256