Ta có:\(a^4;b^4;c^4;d^4\ge0;\forall a;b;c;d\)

Áp dụng BĐT AM-GM, ta có:

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{a^4b^4c^4d^4}\)

\(a^4+b^4+c^4+d^4\ge4abcd\) ( đfcm )

\(\left\{{}\begin{matrix}A=\left(a^4+b^4\right)\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left[\dfrac{\left(a+b\right)^2}{2}\right]^2}{2}\ge\dfrac{\left[\dfrac{4ab}{2}\right]^2}{2}\\B=\left(c^4+d^4\right)\ge\left(c^2+d^2\right)^2\ge\dfrac{\left[\dfrac{\left(c+d\right)^2}{2}\right]^2}{2}\ge\dfrac{\left[\dfrac{4cd}{2}\right]^2}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}A\ge\dfrac{\left(2ab\right)^2}{2}\\B\ge\dfrac{\left(2cd\right)^2}{2}\end{matrix}\right.\)(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 \(\left\{{}\begin{matrix}a^4=b^4\\c^4=d^4\\\left(ab\right)^2=\left(cd\right)^2\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}a=b\\c=d\\ab=cd\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)

giá mà chi a,b,c,d dương

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{a^4b^4c^4d^4}=4\left|abcd\right|\ge4abcd\)

Dấu "=" xảy ra nên: \(a=b=c=d\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\left(1+1\right)=16\)

Với a,b,c,d >0\(a^4+b^4+c^4+d^4=4abcd\Leftrightarrow a^4+b^4+c^4+d^4-4abcd=0\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^4-2c^2d^2+d^4\right)+\left(2a^2b^2+2c^2d^2-4abcd\right)=0\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2-2\left(a^2b^2-2abcd+c^2d^2\right)=0\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2-2\left(ab-cd\right)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a^2-b^2\right)^2\ge0\forall a,b\\\left(c^2-d^2\right)^2\ge0\forall c,d\\\left(ab-cd\right)^2\ge0\forall a,b,c,d\end{matrix}\right.\)

Do đó: \(\left\{{}\begin{matrix}a^2-b^2=0\\c^2-d^2=0\\ab-cd=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2\\c^2=d^2\\ab=cd\end{matrix}\right.\Leftrightarrow a=b=c=d\left(\text{đ}pcm\right)\)

Áp dụng BĐT Cauchy cho 4 số dương:

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{\left(abcd\right)^4}=4abcd\)

(Dấu "="\(\Leftrightarrow a=b=c=d\))

\(\Rightarrow a=b=c=d=\frac{2016}{4}=504\)

Bài này em làm nhầm rồi nhé: chú ý: \(\sqrt[4]{\left(abcd\right)^4}=\left|abcd\right|\ne abcd\)  nhé!

Ta áp dụng Cauchy 2 số

\(a^4+b^4+c^4+d^4\ge2a^2b^2+2c^2d^2\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge2\left(a^2b^2+c^2d^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge2\cdot2abcd\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge4abcd\)

Dấu = khi \(\begin{cases}a^4=b^4\\c^4=d^4\\a^2b^2=c^2d^2\end{cases}\)\(\Rightarrow a=b=c=d\)

Nhanh hơn có thể dùng Cauchy 4 số 

\(a^4+b^4+c^4+d^4\ge4\cdot\sqrt[4]{a^4b^4c^4d^4}\)

\(\Leftrightarrow a^4+b^4+c^4+d^4\ge4abcd\)

Dấu = khi các biến bằng nhau

\(\Leftrightarrow a=b=c=d\)

Áp dụng BĐT Cauchy, ta có:

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{a^4b^4c^4d^4}=4abcd\)

Dấu = xảy ra khi a=b=c=d

Vậy a=b=c=d

a4+b4+c4+d2>4abed