if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$

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.

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!