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Prove that $\sqrt{a^2 + b^2} \geq \sqrt[3]{a^3+b^3}$ for all real $a,b$. I tried with polar coordinates such that: $\sqrt{r^2\sin^2\theta+r^2\cos^2\theta}\geq\sqrt[3]{r^3\sin^3\theta+r^3\cos^3\theta}$ And from here it sort of solves itself, but I'd like to see how to do it without polar coordinates.

Júlio Cezar
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2 Answers2

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Hint: Raise it to 6 power:

$$(a^2+b^2)^3\geq (a^3+b^3)^2$$

then we have to prove $$3a^4b^2+3a^2b^4\geq 2a^3b^3$$

So we have to prove $$3a^2+3b^2 \geq 2ab$$ which is true since $a^2+b^2\geq 2ab$

nonuser
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    As the problem is for 'all real $a$, $b$', strictly speaking we should first reduce the problem to positive $a$, $b$. (Which is trivial, but thought I'd mention it for completeness.) –  Oct 15 '19 at 20:09
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This method has the advantage of having to compute less variables.

Divide by $a$

$$ (1+(b/a)^2)^{1/2} \geq (1+(b/a)^3)^{1/3} $$

Define $t=b/a$. Power 6 both sides will eventually give

$$ (1-t^2)^3 \geq (1+t^3)^2\\ t^2(3t^2-2t+3) \geq 0 $$

Now $t^2 \geq 0$, and you can check the discriminant of the quadratic equation is less than 0.

user651267
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