<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] Your proof is wrong in the first step: $$a^2c^3+b^2d^3\geq a^5c^5+b^5d^5.$$
By Holder $$1>(a^5+b^5)^2(c^5+d^5)^3\geq(a^2c^3+b^2d^3)^5.$$
We can prove the last inequality by the following way.
If $abcd=0$, so our inequality is obvious.
Let $abcd\neq0$, $a=xb$ and $c=yd$.
Thus, we need to prove that $$(x^5+1)^2(y^5+1)^3\geq(x^2y^3+1)^5$$ or $f(x)\geq0,$ where $$f(x)=2\ln(x^5+1)+3\ln(y^5+1)-5\ln(x^2y^3+1).$$
But $$f'(x)=\frac{10x^4}{x^5+1}-\frac{10xy^3}{x^2y^3+1}=\frac{10x(x^3-y^3)}{(x^5+1)(x^2y^3+1)},$$
which gives $x_{min}=y$, $$f(x)\geq f(y)=0$$ and we are done!
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