Proving $\frac14(a+b+c+d)\geq \sqrt[4]{abcd}$

Question

Can somebody help me how to prove that:

$$\frac14(a+b+c+d)\geq \sqrt[4]{abcd}$$

I'm sure it's easy, but I just can't figure out how! I've tried many ways.

Answer 1

Hint: it is the AM-GM inequality for $n=4$: $$x_1,x_2,x_3,x_4\geq 0$$ $$\frac{x_1+x_2+x_3+x_4}{4}\geq \sqrt[4]{x_1x_2x_3x_4}$$ If we have proved the case $n=2$ we can prove this: $$\frac{\frac{x_1+x_2}{2}+\frac{x_3+x_4}{2}}{2}\geq\sqrt{\frac{x_1+x_2}{2}\frac{x_3+x_4}{2}}\geq \sqrt{\sqrt{x_1x_2}\sqrt{x_3x_4}}$$

Answer 2

Use the AM-GM inequality.

This gives ;

$$\frac{a+b+c+d}4\ge\sqrt[4]{abcd}$$

$$\implies a+b+c+d \ge 4\sqrt[4]{abcd}$$

NOTE: As Dr. Sonnhard Graubner pointed out the for cases of $n\ge 3$ the inequality needs to be proved.

As pointed out in his answer , when the case of $n = 2$ is proved;

one can prove ;

$$\frac{\frac{x_1+x_2}{2}+\frac{x_3+x_4}{2}}{2}\geq\sqrt{\frac{x_1+x_2}{2}\frac{x_3+x_4}{2}}\geq \sqrt{\sqrt{x_1x_2}\sqrt{x_3x_4}}$$