`a^4+b^4+c^4+d^4=4abcd`

`<=>a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4=4abcd-2a^2b^2-2c^2d^2`

`<=>(a^2-b^2)^2+(c^2-d^2)^2+2(a^2b^2-2abcd+c^2d^2)>=0`

`<=>(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2=0`

Vì `VT>=0AA a,b,c,d`

Dấu "=" xảy ra khi `a^2=b^2,c^2=d^2,ab=cd`

`<=>a=b=c=d`

áp dụng BDT AM-GM

\(=>a^4+b^4\ge2\sqrt{\left(ab\right)^4}=2a^2b^2\left(1\right)\)

\(=>c^4+d^4\ge2\sqrt{c^4d^4}=2c^2d^2\left(2\right)\)

(1)(2)\(=>a^4+b^4+c^4+d^4\ge2\left(a^2b^2+c^2d^2\right)\ge4abcd\)

dấu"=" xảy ra\(< =>\left\{{}\begin{matrix}a^4=b^4\\c^4=d^4\end{matrix}\right.< =>a=b=c=d}\)

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

\(\left\{{}\begin{matrix}A\ge\dfrac{\left(2ab\right)^2}{2}\\B\ge\dfrac{\left(2cd\right)^2}{2}\end{matrix}\right.\)(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

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 )

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

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

giả sử a=b=c=d => \(a^4+a^4+a^4+a^4=4.a.a.a.a\Leftrightarrow4a^4=4a^4\)=> thỏa mãn điều kiện đầu bài

=> điểu giả sử đúng

Áp đụng BĐT co si ta có:

a⁴+b⁴>2a²b²

b⁴+c⁴>2b²c²

c⁴+d⁴>2c²d²

d⁴+a⁴>2a²d²

=>2(a⁴+b⁴+c⁴+d⁴)>2(a²b²+b²c²+c²d²+a²d²)

=>a⁴+b⁴+c⁴+d⁴>a²b²+b²c²+c²d²+a²d²(1)

Dấu"=" xảy ra <=>a=b=c=d

Tiếp tục ta có:

a²b²+c²d²>2abcd

b²c²+a²d²>2bcd

=>a²b²+b²c²+c²d²+a²d²>4abcd(2)

Từ 1 và 2 =>a⁴+b⁴+c⁴+d⁴>4abcd

Dấu "=" xảy ra <=>a=b=c=d

=>a⁴+b⁴+c⁴+d⁴=4abcd<=>a=b=c=d

1 dòng thôi bạn

Tuy đề bài k cho \(a;b;c;d\) dương nhưng \(a^4;b^4;c^4;d^4\) chắc chắn dương

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

áp dụng BĐT cô si cho 4 số ko âm

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

<=> \(a^4+b^4+c^4+d^4\ge4abcd\) (đpcm)

Cách lớp 8 .

Ta có :

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

\(\Leftrightarrow a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2c^2d^2\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2+2\left(ab-cd\right)^2\ge0\) ( Đúng )

Dấu \("="\) xảy ra khi \(a=b=c=d\)