\(\left(a^2+b+c+d\right)\left(1+b+c+d\right)\ge\left(a+b+c+d\right)^2=16\)
\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}=\dfrac{5-a}{16}\)
Tương tự: \(\dfrac{1}{b^2+c+d+a}\le\dfrac{5-b}{16}\) ...
Cộng vế:
\(P\le\dfrac{20-\left(a+b+c+d\right)}{16}=1\)
Dấu "=" xảy ra khi \(a=b=c=d=1\)