Cho `a,b,c ∈ (1;2)` . Chứng minh rằng : `(bsqrta)/(4bsqrtc – csqrta) + (csqrtb)/(4csqrta – asqrtb) + (asqrtc)/(4asqrtb – bsqrtc) ≥ 1`

Ta có: `a;b;c\in (1;2)`

`=>1<a;b;c<2`

`=>4b>4; ac<4=>\sqrt{ac}<2`

`=>4b-\sqrt{ac}>0`

$\\$

`=> 4b\sqrt{c}-c\sqrt{a}`

`= \sqrt{c}. (4b-\sqrt{ac})>0`

$\\$

Với mọi `a;b;c\in (1;2)`

`=>`$\begin{cases}a+b\ge 2\sqrt{ab}\\b+c\ge 2\sqrt{bc}\end{cases}\ (BĐT\ Cosi)$

`=>(a+b)(b+c)\ge 2\sqrt{ab}.2\sqrt{bc}=4b\sqrt{ac}`

`=>ab+b^2+ac+bc\ge 4b\sqrt{ac}`

`=>b(a+b+c)\ge 4b\sqrt{ac}-ac`

`=>\sqrt{a}b(a+b+c)\ge 4ab\sqrt{c}-ac\sqrt{a}`

`=>b\sqrt{a}.(a+b+c)\ge a(4b\sqrt{c}-c\sqrt{a})`

`=>{b\sqrt{a}}/{4b\sqrt{c}-c\sqrt{a}}\ge a/{a+b+c}`

Tương tự chứng minh được:

`\qquad {c\sqrt{b}}/{4c\sqrt{a}-a\sqrt{b}}\ge b/{a+b+c}`

`\qquad {a\sqrt{c}}/{4a\sqrt{b}-b\sqrt{c}}\ge c/{a+b+c}`

`=>{b\sqrt{a}}/{4b\sqrt{c}-c\sqrt{a}}+{c\sqrt{b}}/{4c\sqrt{a}-a\sqrt{b}}+{a\sqrt{c}}/{4a\sqrt{b}-b\sqrt{c}}\ge {a+b+c}/{a+b+c}=1`

`=>Đpcm`

Dấu “=” xảy ra khi `a=b=c`