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bạn viết sai đề rồi nhé đề đúng là căn(b^2+1/c^2) và căn (c^2 + 1/a^2) ở vế trái chứ ?
Áp dụng BĐT Cô - si, ta có :
\(\left(1.a+\frac{9}{4}.\frac{1}{b}\right)^2\le\left(1^2+\frac{81}{16}\right)\left(a^2+\frac{1}{b^2}\right)\)
\(\Rightarrow\sqrt{a^2+\frac{1}{b^2}}\ge\frac{4}{\sqrt{97}}\left(a+\frac{9}{4b}\right)\).Chứng minh tương tự, ta có:
\(\sqrt{b^2+\frac{1}{c^2}}\ge\frac{4}{\sqrt{97}}\left(b+\frac{9}{4c}\right)\)
\(\sqrt{c^2+\frac{1}{a^2}}\ge\frac{4}{\sqrt{97}}\left(c+\frac{4}{9a}\right)\)
Cộng 3 vế BĐT => đpcm
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow N< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
\(N< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)
\(\frac{1}{2}+\frac{1}{2}cosx=\frac{1}{2}\left(1+cosx\right)=\frac{1}{2}\left(1+2cos^2\frac{x}{2}-1\right)=cos^2\frac{x}{2}\)
Do \(0< x< \frac{\pi}{2}\Rightarrow cos\frac{x}{k}>0\) \(\forall k\) nguyên dương
\(\Rightarrow A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosx}}}\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)
\(A=cos\frac{x}{8}\)
\(\Rightarrow\) Với \(n=\pm8\) thì đẳng thức luôn đúng
\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)
\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)
\(=cos\frac{a}{8}\Rightarrow n=8\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{1}{2\sqrt{2}}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\)
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{1}{2\sqrt{2}}\left(\sqrt{2}.\sqrt{a^2+b^2}+\sqrt{2}.\sqrt{b^2+c^2}+\sqrt{2}.\sqrt{c^2+a^2}\right)\)
\(VT\ge\sqrt{2}.\frac{9}{2\left(a+b+c\right)}\ge\sqrt{2}.\frac{9}{2\sqrt{3\left(a^2+b^2+c^2\right)}}=\frac{3\sqrt{2}}{2}\left(1\right)\)
\(VP\le\frac{1}{2\sqrt{2}}.\frac{2\left(a^2+b^2+c^2\right)+6}{2}=\frac{3\sqrt{2}}{2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow VT\ge VP\)
Dấu \("="\) xảy ra khi \(a=b=c=1\)
\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\)
\(A=2\left(\frac{1}{\sqrt{1}+\sqrt{1}}+\frac{1}{\sqrt{2}+\sqrt{2}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}\right)\)
\(A>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)
\(A>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\right)\)
\(A>2\left(\sqrt{n+1}-1\right)\)