<s>[INST] if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ given $a,b,c,d$ are non  negative real numbers
My try:
It is easy to deduce that $a,b,c,d<1$ ,thus
$$a^2c^3+b^2d^3<a^5c^5+b^5d^5<(a^5+b^5)(c^5+d^5)<1$$
I want to know if there is a more cleaner method to solve this problem and if possible with calculus. I would also  want to know if my proof has a mistake anywhere.
 [/INST] Using the AM-GM inequality, we have
$$2a^5 + 3c^5 \geqslant 5\sqrt[5]{(a^5)^2 \cdot (c^5)^3} = 5a^2 c^3,$$
similar
$$ 2b^5 + 3d^5 \geqslant 5b^2 d^3.$$
Therefore
$$5(a^2c^3+b^2d^3) \leqslant 2(a^5+b^5)+3(c^5+d^5).$$
But $$2(a^5+b^5)+3(c^5+d^5) < 2 \cdot 1 +3\cdot 1 = 5.$$
So
$$a^2c^3+b^2d^3 \leqslant 1.$$
Done.
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