http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/128611/mot-ket-qua-dep.
Áp dụng bài toán trên,chứng minh các bài toán sau đây:
$1.\sqrt{\frac{a^2}{a^2+7ab+b^2}}+\sqrt{\frac{b^2}{b^2+7bc+c^2}}+\sqrt{\frac{c^2}{c^2+7ac+a^2}}\geq 1$(Với mọi a,b,c dương)
2.$\frac{1}{3a^2+(a-1)^2}+\frac{1}{3b^2+(b-1)^2}+\frac{1}{3c^2+(c-1)^2}\geq 1$(a,b,c>0,abc=1)
3.$\sqrt{\frac{a^2}{a^2+\frac{1}{4}bc+c^2}}+\sqrt{\frac{b^2}{b^2+\frac{1}{4}bc+c^2}}+\sqrt{\frac{c^2}{c^2+\frac{1}{4}ac+c^2}}\leq 2$(Với mọi a,b,c dương)
4.$\frac{1}{(a+b)^3}+\frac{1}{(b+c)^3}+\frac{1}{(a+c)^3}\geq \frac{3}{8}$(abc=1,a,b,c>0)
Cần chứng minh $3x^2+(x-1)^2 \le x^4+x^2+1\Leftrightarrow x(x^3-3x+2) \ge 0$
$\Leftrightarrow x^3-3x+2 \ge 0 $(đúng theo bđt $AM-GM : x^3+1+1-3x \ge 3x-3x=0$)
1)
Đặt $\frac ba =x, \frac cb =y, \frac ac =z (xyz=1)$
Có $VT=\sum \sqrt{\frac{a^2}{a^2+7ab+b^2}}= \sum\frac 1{\sqrt{1+7\frac ba+(\frac ba)^2}}= \sum\frac 1{\sqrt{x^2+7x+1}}$
Ta cần chứng minh $\frac 1{\sqrt{x^2+7x+1}} \ge \frac 1{x+\sqrt x+1} (1)$ đúng từ đó áp dụng bài toán trên suy ra đpcm
Thật vậy $(1)\Leftrightarrow (x+\sqrt x+1)^2 \ge x^2+7x+1\Leftrightarrow (\sqrt{x}-1)^2 \ge 0$ (luôn đúng)

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