Question 85
Prove that
$$
64 \frac{a b c d+1}{(a+b+c+d)^{2}} \leq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}
$$
for $a, b, c, d\gt 0$.
$$
64 \frac{a b c d+1}{(a+b+c+d)^{2}} \leq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}
$$
for $a, b, c, d\gt 0$.