Ta có : \(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{2}{3}\)

\(\Leftrightarrow3\left(a+b+c+d\right)^2\ge8\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)-2\left(ab+ac+ad+bc+bd+cd\right)\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)+\left(c^2-2cd+d^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\) (luôn đúng)

Vậy bđt ban đầu được chứng minh

(a+b+c+d)2\(\ge\frac{8}{3}\)(ab+ac+ad+bc+bd+cd)

<=>(a+b)²+2(a+b)(c+d)+(c+d)²\(\ge\).....

<=>a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd)\(\ge\)....

<=>3a²+3b²+3c²+3d²+6(ab+ac+ad+bc+bd+cd)\(\ge\)8(ab+ac+ad+bc+bd+cd)

<=> 3a²+3b²+3c²+3d²-2ab -2ac-2bc-2ad-2bd-2cd\(\ge\)0

<=> (a²-2ab+b²)+(a²-ac+c²)+(a²-2ad+d²)+(b²-2bc+c²)+(b²-2bd+d²)+(c²-2cd+d²)>=0

<=> (a-b)²+(a-c)²+(a-d)²+(b-c)²+(b-d)²+(c-d)²>=0 (DPCM)

Dau ''='' xay ra khi a=b=c=d

chắc áp dụng định lý Lagrange và bất đẳng thức AM-GM

Ta có :

 \(3\left(a^2+b^2+c^2+d^2\right)-2\left(ab+ac+ad+bc+bd+cd\right)\)

\(=\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\)

\(\Rightarrow a^2+b^2+c^2+d^2\ge\frac{2}{3}\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Rightarrow\left(a+b+c+d\right)^2=a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)\)

\(\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\left(đpcm\right)\)

\(\left(a+b+c+d\right)^2\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)\)\(+\left(c^2-2cd+d^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\) ( đúng )
=> Đpcm

Nhìn BĐT 4 số ngán quá

\(1\ge4\sqrt[4]{\frac{1}{a^2b^2c^2d^2}}\Rightarrow abcd\ge16\)

\(\Rightarrow VT=\frac{abcd}{8}+2\ge4\) (1)

Mà \(VP=\frac{a+c}{\sqrt{ac}}+\frac{b+d}{\sqrt{bd}}\le\frac{2\left(a+c\right)}{a+c}+\frac{2\left(b+d\right)}{b+d}=4\) (2)

(1);(2) \(\Rightarrow\) đpcm

Dấu "=" xảy ra khi \(a=b=c=d=2\)

Akai Haruma

@Nguyễn Việt Lâm anh giúp em vs !!!

\(a+b+c=6abc\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=6\)

Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow xy+xz+yz=6\)

\(P=\sum\frac{\frac{1}{yz}}{\frac{1}{x^3}\left(\frac{1}{z}+\frac{2}{y}\right)}=\sum\frac{x^3}{y+2z}=\sum\frac{x^4}{xy+2xz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+xz+yz\right)}\ge\frac{\left(xy+xz+yz\right)^2}{3\left(xy+xz+yz\right)}=2\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{2}\Leftrightarrow a=b=c=\frac{1}{\sqrt{2}}\)