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![](https://rs.olm.vn/images/avt/0.png?1311)
\(sigma\frac{a}{1+b-a}=sigma\frac{a^2}{a+ab-a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\frac{1}{b^2+c^2}=\frac{1}{1-a^2}=1+\frac{a^2}{b^2+c^2}\le1+\frac{a^2}{2bc}\)
Tương tự cộng lại quy đồng ta có đpcm
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Sigma\frac{b+1}{8-\sqrt{a}}\le\Sigma\frac{2\left(b+1\right)}{15-a}=\Sigma\frac{2\left(a+2b+c\right)}{4a+5b+5c}\)(AM-gm)
Đặt \(\left\{\begin{matrix}x=4a+5b+5c\\y=4b+5a+5c\\z=4c+5a+5b\end{matrix}\right.\)suy ra...
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Đề thiếu
2. BĐT cần chứng minh tương đương:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Ta có:
\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)
3.
Ta có:
\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)
\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)
\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)
Lại có:
\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)
4.
Ta có:
\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)
\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
5.
Ta có:
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)
\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(a+b+\sqrt{2\left(a+c\right)}=a+b+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(a+c\right)}{2}}\)
Hoàn toàn tương tự ta có:
\(\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}\le\frac{2}{27\left(b+c\right)\left(b+a\right)}\);
\(\frac{1}{\left(c+b+\sqrt{\left(c+b\right)}\right)^3}\le\frac{2}{27\left(c+a\right)\left(c+b\right)}\)
Cộng theo bất đẳng thức trên ta được:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{4\left(a+b+c\right)}{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Do đó:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{1}{6\left(ab+bc+ca\right)}\)
Vậy bất đẳng thức được chứng minh, bất đẳng thức xày ra khi \(a=b=c=\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho bài toán phụ : Cho a ; b là các số thực dương
C/m : \(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{ab+1}\)
Do a ; b là các số thực dương \(\Rightarrow ab\ge1\)
Ta có : \(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{ab+1}\)
\(\Leftrightarrow\frac{1}{a^2+1}-\frac{1}{ab+1}+\frac{1}{b^2+1}-\frac{1}{ab+1}\ge0\)
\(\Leftrightarrow\frac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\frac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{ab^3-a^2b^2+ab-a^2+a^3b-a^2b^2+ab-b^2}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{ab\left(a^2+b^2\right)+2ab-2a^2b^2-a^2-b^2}{...}\ge0\)
\(\Leftrightarrow\frac{\left(a^2+b^2\right)\left(ab-1\right)-2ab\left(ab-1\right)}{...}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{...}\ge0\)
Dễ thấy mẫu luôn dương , tử \(\ge0\) => luôn đúng
=> BĐT được c/m
Áp dụng BĐT phụ ( từ bài toán phụ trên ) , ta có :
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{2}{ab+1}+\frac{1}{c^2+1}=\frac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\frac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)
( * )
Có : \(\frac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}-\frac{3}{2}=\frac{4c^2+2ab+6-3abc^2-3c^2-3ab-3}{...}=\frac{c^2+3-ab-3abc^2}{...}=\frac{c^2+bc+ac-3abc^2}{...}=\frac{c\left(a+b+c-3abc\right)}{...}\)\(\left(ab+bc+ac=3\right)\) ( 1 )
Do a , b , c là các số thực dương , áp dụng BĐT Cô - si cho 3 số , ta có : \(\left(a+b+c\right)\left(ab+bc+ac\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{a^2b^2c^2}=9abc\)
\(\Rightarrow a+b+c\ge3abc\left(ab+bc+ac=3\right)\) ( 2 )
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\frac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}-\frac{3}{2}\ge0\)
\(\Rightarrow\frac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\frac{3}{2}\) ( *' )
Từ (*) và (*') => ĐPCM
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
h ms trả lời :(
Thì sao