Question

Cho a, b, c là ba số thực không âm thỏa mãn \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)

Chứng minh rằng: \(\frac{\sqrt{a}}{1+a}+\frac{\sqrt{b}}{1+b}+\frac{\sqrt{c}}{1+c}=\frac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

GIÚP MÌNH VỚI!!!! QAQ

Answer 1

Đặt $(\sqrt{a}, \sqrt{b}, \sqrt{c})=(x,y,z)$ thì bài toán trở thành:
Cho $x,y,z$ không âm thỏa mãn: $x^2+y^2+z^2=x+y+z=2$

Chứng minh rằng:

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\)

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Lời giải:

$xy+yz+xz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{2^2-2}{2}=1$. Do đó:

$\text{VT}=\frac{x}{x^2+xy+yz+xz}+\frac{y}{y^2+xy+yz+xz}+\frac{z}{z^2+xy+yz+xz}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}{(z+x)(z+y)}$

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

$\text{VP}=\frac{2}{\sqrt{(x^2+xy+yz+xz)(y^2+xy+yz+xz)(z^2+xy+yz+xz)}}$

$=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{(x+y)(y+z)(x+z)}$

Do đó: $\text{VT}=\text{VP}$ (đpcm)

Answer 2

Đặt $(\sqrt{a}, \sqrt{b}, \sqrt{c})=(x,y,z)$ thì bài toán trở thành:
Cho $x,y,z$ không âm thỏa mãn: $x^2+y^2+z^2=x+y+z=2$

Chứng minh rằng:

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\)

------------------------------------

Lời giải:

$xy+yz+xz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{2^2-2}{2}=1$. Do đó:

$\text{VT}=\frac{x}{x^2+xy+yz+xz}+\frac{y}{y^2+xy+yz+xz}+\frac{z}{z^2+xy+yz+xz}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}{(z+x)(z+y)}$

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

$\text{VP}=\frac{2}{\sqrt{(x^2+xy+yz+xz)(y^2+xy+yz+xz)(z^2+xy+yz+xz)}}$

$=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{(x+y)(y+z)(x+z)}$

Do đó: $\text{VT}=\text{VP}$ (đpcm)