Question

help 

Answer 1

Bài 62:

Ta có:

$\text{VT}=\sum \frac{a}{a+bc}=\sum \frac{a}{a(a+b+c)+bc}=\sum \frac{a}{(a+b)(a+c)}$

$=\frac{a(b+c)+b(a+c)+c(a+b)}{(a+b)(b+c)(c+a)}=\frac{2(ab+bc+ac)}{(a+b)(b+c)(c+a)}$
$=\frac{2(ab+bc+ac)}{(a+b+c)(ab+bc+ac)-abc}$

$=\frac{2(ab+bc+ac)}{ab+bc+ac-abc}$

Ta sẽ cm $\frac{2(ab+bc+ac)}{ab+bc+ac-abc}\leq \frac{9}{4}$

$\Leftrightarrow 8(ab+bc+ac)\leq 9(ab+bc+ac-abc)$

$\Leftrightarrow 9abc\leq ab+bc+ac$

BĐT luôn đúng do theo BĐT AM-GM thì:

$ab+bc+ac=(ab+bc+ac)(a+b+c)\geq 3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}=9abc$

Vậy ta có ddpcm.

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Answer 2

Bài 63:

$\text{VT}=\sum \frac{x}{(2x+y+z)^2}=\sum \frac{x}{[(x+y)+(x+z)]^2}\leq \sum \frac{x}{4(x+y)(x+z)}$ (áp dụng BĐT Cô-si)

$=\frac{1}{4}\sum \frac{x}{(x+y)(x+z)}=\frac{1}{4}.\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}$

$=\frac{1}{2}.\frac{xy+yz+xz}{(x+y)(y+z)(x+z)}$

$=\frac{1}{2}.\frac{xy+yz+xz}{(x+y+z)(xy+yz+xz)-xyz}$

$\leq \frac{1}{2}.\frac{xy+yz+xz}{(x+y+z)(xy+yz+xz)-\frac{(x+y+z)(xy+yz+xz)}{9}}$ (áp  dụng BĐT Cô-si)

$=\frac{1}{2}.\frac{xy+yz+xz}{\frac{8}{9}(x+y+z)(xy+yz+xz)}=\frac{9}{16}.\frac{xy+yz+xz}{(x+y+z)(xy+yz+xz)}=\frac{9}{16}.\frac{xy+yz+xz}{3(xy+yz+xz)}=\frac{3}{16}$ 

Ta có đpcm

Dấu "=" xảy ra khi $x=y=z=1$