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![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn ơi đề bài có điều kiện a, b, c không vậy. Hay là a, b, c bất kì?
![](https://rs.olm.vn/images/avt/0.png?1311)
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(3=ab+bc+ac\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow1\ge abc\)
\(\frac{bc}{a^2\left(b+2c\right)}+\frac{ac}{b^2\left(c+2a\right)}+\frac{ab}{c^2\left(a+2b\right)}\)
\(=\frac{\left(bc\right)^2}{abc\left(ab+2ac\right)}+\frac{\left(ac\right)^2}{abc\left(bc+2ab\right)}+\frac{\left(ab\right)^2}{abc\left(ca+2cb\right)}\)
\(\ge\frac{\left(ab+bc+ac\right)^2}{abc\left(3ab+3ac+3bc\right)}\)\(=\frac{3^2}{9abc}\)\(\ge1\)\(\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=a-\frac{ac}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}$
$=\sum a-\sum \frac{ac}{c+a^2}$
Áp dụng BĐT AM-GM: $c+a^2\geq 2a\sqrt{c}$
$\Rightarrow A\geq \sum a-\frac{1}{2}\sum \sqrt{c}$
Áp dụng BĐT Cauchy-Schwarz:
$(\sum \sqrt{c})^2\leq (c+a+b)(1+1+1)$
$\Rightarrow \sum \sqrt{c}\leq 3\sum a$
Do đó $A\geq \sum a-\frac{1}{2}\sqrt{3\sum a}$
Đặt $\sqrt{3\sum a}=t$ thì $A\geq \frac{t^2}{3}-\frac{t}{2}(*)$
Từ điều kiện $ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Áp dụng BĐT Cauchy-Schwarz:
$3=\sum \frac{1}{a}\geq \frac{9}{\sum a}\Rightarrow \sum a\geq 3$
$\Rightarrow t=\sqrt{3\sum a}\geq 3$
Do đó:
$\frac{t^2}{3}-\frac{t}{2}=(t-3)(\frac{t}{3}+\frac{1}{2})+\frac{3}{2}\geq \frac{3}{2}$ với mọi $t\geq 3(**)$
Từ $(*); (**)\Rightarrow A\geq \frac{3}{2}$
Vậy $A_{\min}=\frac{3}{2}$ khi $a=b=c=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(2a^2+b^2\right)=\left(1^2+1^2+1^2\right)\left(a^2+a^2+b^2\right)\ge\left(a+a+b\right)^2=\left(2a+b\right)^2\)
\(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(gt\rightarrow7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\)
\(\Leftrightarrow7\left(x+y+z\right)^2=20\left(xy+yz+zx\right)+2015\)
Ta có: \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)
\(\Rightarrow7\left(x+y+z\right)^2\le\frac{20}{3}\left(x+y+z\right)^2+2015\)
\(\Leftrightarrow\frac{1}{3}\left(x+y+z\right)^2\le2015\)
\(\Leftrightarrow x+y+z\le\sqrt{6045}\)
\(P\le\frac{1}{3}\left(x+y+z\right)\le\frac{\sqrt{6045}}{3}\)
Dấu bằng xảy ra khi \(x=y=z=\frac{\sqrt{6045}}{3}\)hay \(a=b=c=\left(\frac{\sqrt{6045}}{3}\right)^{-1}\)