Ta có : \frac{a}{b+2c+3d} = \frac{a^2}{ab+2ac+3ad}Tương tự ..........cộng lại \frac{a}{b+2c+3d} + \frac{b}{c+2d+3a} + \frac{c}{d+2a+3b} + \frac{d}{a+2b+3c} = \frac{a^2}{ab+2ac+3ad} + \frac{b^2}{bc+2db+3ab} + .............>= \frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}Ta có \frac{3}{2}. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) = \frac{3}{2}.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}=\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2} >= 0 \Rightarrow 4(ab+bc+ca+ad+bd+cd) <= \frac{3}{2} .(a+b+c+d)^2\Rightarrow dpcm
Ta có :
$\frac{a}{b+2c+3d}
$ =
$\frac{a^2}{ab+2ac+3ad}
$Tương tự ..........cộng lại
$\frac{a}{b+2c+3d}
$ +
$\frac{b}{c+2d+3a}
$ +
$\frac{c}{d+2a+3b}
$ +
$\frac{d}{a+2b+3c}
$ =
$\frac{a^2}{ab+2ac+3ad}
$ +
$\frac{b^2}{bc+2db+3ab}
$ + .............>=
$\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd}
$Ta có
$\frac{3}{2}
$. (a+b+c+d)^2 - 4(ab+bc+ca+ad+bd+cd) =
$\frac{3}{2}
$.(a^2+b^2+c^2+d^2+2ab+2bc+2cd+2da+2ac+2bd)-4(ab+bc+ca+ad+bd+cd)=
$\frac{3(a^2+b^2+c^2+d^2)-2(ab+bc+ca+ad+bd+cd)}{2}
$=
$\frac{(a-b)^2+(b-c)^2+(c-a)^2+(a-d)^2+(b-d)^2}{2}
$ $\g
eq $0
$\Rightarrow
$ 4(ab+bc+ca+ad+bd+cd)
$\l
eq $ $\frac{3}{2}
$ .(a+b+c+d)^2
$\Rightarrow
$ dpcm