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In an exercise of a limit of a function of two variables in the solution I read this inequality:

$$ \frac{x^2y^2}{(x^2+y^2)^\frac{3}{2}} \le \frac{1}{2}\sqrt{x^2+y^2} $$

how did they arrive at this result?

Salmon
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    Have you tried manipulating this inequality? It's not a standard inequality, but it's pretty easy to verify it via algebra - if you have some questions about how to do that, it might be good to add that as context to this question so that we could give a more useful answer than just some random piece of algebra. – Milo Brandt Dec 25 '20 at 22:43
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    Since $4x^2 y^2\leq (x^2+y^2)^2$, $$\frac{x^2 y^2}{(x^2+y^2)^{3/2}}\leq\frac{1}{\color{red}4}\sqrt{x^2+y^2}$$ actually holds. – Jack D'Aurizio Dec 25 '20 at 22:46

2 Answers2

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That inequality is equivalent to$$x^2y^2\leqslant\frac12\sqrt{x^2+y^2}(x^2+y^2)^{3/2}=\frac{(x^2+y^2)^2}2.$$Besides,$$(x^2+y^2)^2-2x^2y^2=x^4+y^4\geqslant0.$$

José Carlos Santos
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You can clearly see that the inequality is homogeneous. Now just assume $x^2+y^2=1$ and the inequality is trivial! But as Jack D'Aurizio stated, the RHS constant is 1/4 not 1/2.

Aryan
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