Example 1.15.4. Let \(a, b, c, d\) be positive real numbers such that that
\[2(a+b+c+d)^{2}=5\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\]

Find the minimum value of
\[P=\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d}\]
SOLUTION. First we guess that the equality holds for $b=c=d$ (in this case, we find out $a=2+\sqrt{5}, b=c=d=1$ and permutations) and this is the key to the solution. Indeed, denote $k=(2+\sqrt{5})^{3}+(2+\sqrt{5})^{-1}>78$, we will prove
$$a^{4}+b^{4}+c^{4} \geq 4 a b c d$$

WLOG, assume that $a \geq b \geq d \geq c$. We fix $d$ and suppose that $a+b+c=$ const, $a^{2}+$ $b^{2}+c^{2}=$ const. By hypothesis, we have
$$4(a+b+c+d)^{2}=10\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geq 5\left((a+b)^{2}+(c+d)^{2}\right)$$
so
$$(a+b)^{2}+(c+d)^{2} \leq 8(a+b)(c+d) \Rightarrow a+b \leq(4+\sqrt{15})(c+d)$$
and therefore
$$a+b<(4+\sqrt{15})(c+d)<(8+2 \sqrt{15}) d<18 d \Rightarrow 4(a+b+c) \leq 76 d$$

Moreover, notice that
$$a^{4}+b^{4}+c^{4}-k d \cdot a b c=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2(a b+b c+c a)^{2}+[4(a+b+c)-k d] a b c$$
is a strictly decreasing function of $a b c$ (because $a+b+c$ and $a^{2}+b^{2}+c^{2}$ have been fixed already). By the second corollary, it suffices to consider the initial problem in case $a \geq b=c=d$. If $b=c=d$, we deduce that $a=(2+\sqrt{5}) b$ by hypothesis, so
$$a^{4}+b^{4}+c^{4}+d^{4}=\left((2+\sqrt{5})^{4}+1\right) b^{4}=k a b c d$$