If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$ 
If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$

$\bf{My\; Try::} $Using $\bf{A.M\geq G.M}$
$$\frac{a^{11}+a^{11}+a^{11}+a^{11}+a^{11}+x^{11}+x^{11}+x^{11}+x^{11}+x^{11}+x^{11}}{11}\geq \sqrt[11]{a^{55}x^{66}}$$
and  $$\frac{b^{11}+b^{11}+b^{11}+b^{11}+b^{11}+y^{11}+y^{11}+y^{11}+y^{11}+y^{11}+y^{11}}{11}\geq \sqrt[11]{b^{55}y^{66}}$$
Now Adding These two inequalities, We get
$$a^{5}x^6+b^5y^6\leq \frac{5(a^{11}+b^{11})+6(x^{11}+y^{11})}{11}\leq \frac{5+6}{11}=1$$
My question is can we solve in some short way, If yes then plz explain me, Thanks

Using Hölder's inequality with $p=\frac{11}{5}$ and $q=\frac{11}{6}$:
\begin{align}a^5x^6+b^5y^6 & \leq ((a^5)^p+(b^5)^p)^{\frac{1}{p}}((x^6)^q+(y^6)^q)^{\frac{1}{q}}\\ & \leq (a^{11}+b^{11})^{\frac{5}{11}}(x^{11}+y^{11})^{\frac{6}{11}}=1.\end{align}
