cho ab>1 .cmr (1/a²+1)+(1/ b²+1)>= 2/ab+1
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Mọi người giúp em với, em cần gấp lắm ạ. Em cảm ơn mọi người nhiều ạ
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1.
\(\left(1+a\right)^2=\left(1.1+\sqrt{\frac{a}{b}}.\sqrt{ab}\right)^2\le\left(1+\frac{a}{b}\right)\left(1+ab\right)=\frac{\left(a+b\right)\left(1+ab\right)}{b}\)
\(\Rightarrow\frac{1}{\left(1+a\right)^2}\ge\frac{b}{\left(a+b\right)\left(1+ab\right)}\)
\(\left(1+b\right)^2\le\frac{\left(a+b\right)\left(1+ab\right)}{a}\Rightarrow\frac{1}{\left(1+b\right)^2}\ge\frac{a}{\left(a+b\right)\left(1+ab\right)}\)
\(\Rightarrow\frac{1}{\left(1+a\right)^2}+\frac{1}{\left(1+b\right)^2}\ge\frac{a}{\left(a+b\right)\left(1+ab\right)}+\frac{b}{\left(a+b\right)\left(1+ab\right)}=\frac{1}{1+ab}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=1\)
2.
\(P=\sqrt{\frac{a^2}{a^4+3}}+\sqrt{\frac{b^2}{b^4+3}}\le\sqrt{2\left(\frac{a^2}{a^4+3}+\frac{b^2}{b^4+3}\right)}\)
Đặt \(\left(a^2;b^2\right)=\left(x;y\right)\Rightarrow xy=1\)
\(Q=\frac{x}{x^2+3}+\frac{y}{y^2+3}=\frac{x}{x^2+3}+\frac{x}{3x^2+1}-\frac{1}{2}+\frac{1}{2}\)
\(Q=\frac{-\left(x-1\right)^2\left(3x^2-2x+3\right)}{2\left(x^2+3\right)\left(3x^2+1\right)}+\frac{1}{2}\le\frac{1}{2}\)
\(\Rightarrow P\le\sqrt{2Q}\le1\)
\(P_{max}=1\) khi \(a=b=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT cô si ta có:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{\sqrt{\left(a^2+1\right)\left(b^2+1\right)}}\)
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{a^2+1+b^2+1}{2}=\frac{a^2+b^2+2}{2}\)
\(\Rightarrow\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{4}{a^2+b^2+2}\left(1\right)\)
Ta thấy: \(\frac{4}{a^2+b^2+2}=1-\frac{a^2+b^2-2}{a^2+b^2+2}\left(2\right)\)
Ta có: \(a^2+b^2+2\ge2ab+2=2\left(ab+1\right)\)
\(\Rightarrow\frac{1}{a^2+b^2+2}\le\frac{1}{2\left(ab+1\right)}\)
\(\Rightarrow\frac{-1}{a^2+b^2+2}\ge\frac{-1}{2\left(ab+1\right)}\)
Mà \(a^2+b^2-2\ge2\left(ab-1\right)\)
\(\Rightarrow1-\frac{a^2+b^2-2}{a^2+b^2+2}\ge1-\frac{2\left(ab-1\right)}{2\left(ab+1\right)}\)
\(=1-\frac{ab-1}{ab+1}=\frac{ab+1-ab+1}{ab+1}=\frac{2}{ab+1}\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\) và \(\left(3\right)\) suy ra:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{ab+1}\) (Đpcm)