Prove the inequality.Let a, b and c be nonnegative real numbers.

Question

Let $a$, $b$ and $c$ be nonnegative real numbers. Prove that $a^4+b^4+c^2\ge 8^{½}abc$

Answer 1

Using AM,GM Inequality

$$a^4+b^4\ge 2a^2b^2$$ which can be demonstrated as $$a^4+b^4=(a^2-b^2)^2+2a^2b^2\ge 2a^2b^2$$ as the square of any real number is $\displaystyle\not<0$

Again similarly, $$2a^2b^2+c^2\ge 2\sqrt{2a^2b^2\cdot c^2}$$

Answer 2

Another way: $$ a^4+b^4+\frac{c^2}{2}+\frac{c^2}{2}\geq 4\left(a^4b^4\frac{c^2}{2}\frac{c^2}{2}\right)^\frac 14=2^\frac 32 abc. $$