\(\left\{{}\begin{matrix}A=\left(a^4+b^4\right)\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left[\dfrac{\left(a+b\right)^2}{2}\right]^2}{2}\ge\dfrac{\left[\dfrac{4ab}{2}\right]^2}{2}\\B=\left(c^4+d^4\right)\ge\left(c^2+d^2\right)^2\ge\dfrac{\left[\dfrac{\left(c+d\right)^2}{2}\right]^2}{2}\ge\dfrac{\left[\dfrac{4cd}{2}\right]^2}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A\ge\dfrac{\left(2ab\right)^2}{2}\\B\ge\dfrac{\left(2cd\right)^2}{2}\end{matrix}\right.\)(1)
\(\left\{{}\begin{matrix}A\ge0\\B\ge0\end{matrix}\right.\)(2)
(1) và (2) \(\Rightarrow A+B\ge\dfrac{\left(2ab\right)^2+\left(2cd\right)^2}{2}\ge\dfrac{2\left(4abcd\right)}{2}=4abcd\)
đẳng thức khi a=b=c=d
Ta có BĐT \(a+b\ge2\sqrt{ab}\Leftrightarrow\left(a+b\right)^2\ge\left(2\sqrt{ab}\right)^2\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\forall a,b\)
Đẳng thức xảy ra khi \(\left(a-b\right)^2=0\Rightarrow a=b\)
Vậy ta có: \(a^4+b^4\ge2\sqrt{a^4b^4}=2a^2b^2\)
\(c^4+d^4\ge2\sqrt{c^4d^4}=2c^2d^2\)
Cộng theo vế 2 BĐT trên ta có:
\(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2=2\left[\left(ab\right)^2+\left(cd\right)^2\right]\)
Lại có: \(\left(ab\right)^2+\left(cd\right)^2\ge2\sqrt{\left(ab\right)^2\left(cd\right)^2}=2abcd\)
\(\Rightarrow2\left[\left(ab\right)^2+\left(cd\right)^2\right]\ge2\cdot2abcd=4abcd\)
\(\Rightarrow VT=a^4+b^4+c^4+d^4\ge4abcd=VP\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a^4=b^4\\c^4=d^4\\\left(ab\right)^2=\left(cd\right)^2\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}a=b\\c=d\\ab=cd\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
