Đặt $\begin{cases}x=\sqrt{a^2+b^2}\\y=\sqrt{b^2+c^2}\\z=\sqrt{a^2+c^2}\end{cases}$ `(x;y;z>0)`
`=>x+y+z=1` vì `\sum\sqrt{a^2+b^2}=1`
`\qquad` $\begin{cases}x^2=a^2+b^2\\y^2=b^2+c^2\\z^2=a^2+c^2\end{cases}$
`=>`$\begin{cases}x^2+z^2-y^2=2a^2\\x^2+y^2-z^2=2b^2\\y^2+z^2-x^2=2c^2\end{cases}$
`=>`$\begin{cases}a^2=\dfrac{x^2+z^2-y^2}{2}\\b^2=\dfrac{x^2+y^2-z^2}{2}\\c^2=\dfrac{y^2+z^2-x^2}{2}\end{cases}$
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Với mọi `a;b;c>0` ta có:
`\qquad (a-b)^2\ge 0`
`=>a^2+b^2\ge 2ab`
`=>2(a^2+b^2)\ge a^2+2ab+b^2`
`=>2(a^2+b^2)\ge (a+b)^2`
`=>2x^2\ge (a+b)^2``=>a+b\le \sqrt{2}x`
`=>1/{a+b}\ge 1/{\sqrt{2}x}`
Tương tự: `1/{b+c}\ge 1/{\sqrt{2}y}; 1/{a+c}\ge 1/{\sqrt{2}z}`
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Bất đẳng thức Bunhiacopxki:
`\qquad (1+1+1)(x^2+y^2+z^2)\ge (1.x+1 y+1.z)^2`
`=>3(x^2+y^2+z^2)\ge (x+y+z)^2`
Bất đẳng thức Cosi với `3` số dương:
`(x+y+z)\ge `$3\sqrt[3]{xyz}$
`(1/x+1/y+1/z)\ge 3`$\sqrt[3]{\dfrac{1}{x} . \dfrac{1}{y} . \dfrac{1}{z}}$
`=>(x+y+z)(1/x+1/y+1/z)\ge 9`
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Đặt `M=a^2/{b+c}+b^2/{a+c}+c^2/{a+b}`
`={x^2+z^2-y^2}/{2(b+c)}+{x^2+y^2-z^2}/{2(a+c)}+{y^2+z^2-x^2}/{2(a+b)}`
`=>M\ge {x^2+z^2-y^2}/{2\sqrt{2}y}+{x^2+y^2-z^2}/{2\sqrt{2}z}+{y^2+z^2-x^2}/{2\sqrt{2}x}`
`\quad \ge 1/{2\sqrt{2}}. ({x^2+z^2-y^2}/y+{x^2+y^2-z^2}/z+{y^2+z^2-x^2}/x)`
`\qquad \ge 1/{2\sqrt{2}}.({x^2+z^2-y^2}/y+2y-2y+{x^2+y^2-z^2}/z+2z-2z+{y^2+z^2-x^2}/x+2x-2x)`
`\ge 1/{2\sqrt{2}}.[{x^2+z^2+y^2}/y+{x^2+y^2+z^2}/z+{y^2+z^2+x^2}/x-2(x+y+z)]`
`\quad \ge 1/{2\sqrt{2}}. [(x^2+y^2+z^2).(1/x+1/y+1/z)-2]`
`\quad \ge 1/{2\sqrt{2}}. [1/ 3 . 3. (x^2+y^2+z^2)(1/x+1/y+1/z)-2]`
`\quad \ge 1/{2\sqrt{2}}.[1/ 3 . (x+y+z)^2(1/x+1/y+1/z)-2]` (BĐT Bunhiacopxki)
`\quad \ge 1/{2\sqrt{2}}. (1/ 3 . (x+y+z)(x+y+z).(1/x+1/y+1/z)-2]`
`\quad \ge 1/{2\sqrt{2}}.(1/ 3 . 1 . 9-2)`
`\quad \ge 1/{2\sqrt{2}}. 1=1/{2\sqrt{2}`
Dấu "=" xảy ra khi:
$\quad \begin{cases}a=b=c\\3\sqrt{a^2+a^2}=1\end{cases}$`=>`$\begin{cases}a=b=c\\2a^2=\dfrac{1}{9}\end{cases}$
`=>a=b=c=1/{3\sqrt{2}}`
Vậy `a^2/{b+c}+b^2/{a+c}+c^2/{a+b}\ge 1/{2\sqrt{2}}` (đpcm)
