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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{1+a^2}+\sqrt{1-a^2}=x\Rightarrow\sqrt{2}\le x\le2\)
\(x^2=2+2\sqrt{1-a^4}\Rightarrow\sqrt{1-a^4}=\dfrac{x^2-2}{2}\)
\(\Rightarrow\dfrac{x^2-2}{2}+\left(b+1\right)x+b-4\le0\)
\(\Rightarrow x^2+2\left(b+1\right)x+2b-10\le0\)
\(\Rightarrow x^2+2x-10\le-2b\left(x+1\right)\)
\(\Rightarrow-2b\ge\dfrac{x^2+2x-10}{x+1}\)
\(\Rightarrow-2b\ge\max\limits_{\left[\sqrt{2};2\right]}f\left(x\right)\) với \(f\left(x\right)=\dfrac{x^2+2x-10}{x+1}\)
Xét trên \(\left[\sqrt{2};2\right]\) ta có:
\(f\left(x\right)=\dfrac{3x^2+6x-30}{3\left(x+1\right)}=\dfrac{3x^2+8x-28-2\left(x+1\right)}{3\left(x+1\right)}=\dfrac{\left(3x+14\right)\left(x-2\right)}{3\left(x+1\right)}-\dfrac{2}{3}\le-\dfrac{2}{3}\)
\(\Rightarrow-2b\ge-\dfrac{2}{3}\Rightarrow b\le\dfrac{1}{3}\)
Vậy \(b_{max}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=\left|x-1\right|=1-x\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=\left|y-1\right|=1-y\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=\left|z-1\right|=1-z\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2\Sigma_{sym}\sqrt{ab}+3\left(a+b+c+d\right)\le6\left(a+b+c+d\right)\le6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
Dễ thấy \(\left(\sqrt{1+a^2}-\sqrt{1-a^2}\right)^2=2-2\sqrt{1-a^4}\) nên đặt \(\sqrt{1+a^2}-\sqrt{1-a^2}=t\) thì
\(GT\Leftrightarrow\frac{2-t^2}{2}+\left(b-1\right)t+b-4\le0\)\(\Leftrightarrow t^2-2\left(b-1\right)t-2b+6\ge0\)
Coi đây là Pt ẩn t , dễ thấy hệ số của \(t^2\)và tam thức bậc 2 ẩn t cùng dấu . Do đó \(\Delta'\le0\)
---> tự giải
cho x,y,z>0 chứng minh rằng
\(\frac{xy}{x^2+yz+zx}+\frac{yz}{y^2+zx+xy}+\frac{zx}{z^2+xy+yz}\le\frac{x^2+y^2+z^2}{xy+xz+zx}\)