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1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
Tham khảo
Câu hỏi của Châu Trần - Toán lớp 9 - Học toán với OnlineMath
Đặt vế trái là P và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=4\)
Ta cần chứng minh: \(P=\frac{1}{xy+2yz+zx}+\frac{1}{xy+yz+2zx}+\frac{1}{2xy+yz+zx}\le\frac{1}{xyz}\)
\(P=\frac{1}{xy+yz+yz+zx}+\frac{1}{xy+yz+zx+zx}+\frac{1}{xy+xy+yz+zx}\)
\(P\le\frac{1}{16}\left(\frac{1}{xy}+\frac{2}{yz}+\frac{1}{zx}+\frac{1}{xy}+\frac{1}{yz}+\frac{2}{zx}+\frac{2}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(P\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{4}\left(\frac{x+y+z}{xyz}\right)=\frac{1}{4}.\frac{4}{xyz}=\frac{1}{xyz}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=\frac{4}{3}\) hay \(a=b=c=\frac{16}{9}\)
Đặt \(A=\frac{a}{\sqrt{a^2+2bc}}+\frac{b}{\sqrt{b^2+2ca}}+\frac{c}{\sqrt{c^2+2ab}}\left(a,b,c>0\right)\)
Ta có:
\(A=\sqrt{1-\frac{2bc}{a^2+2bc}}+\sqrt{1-\frac{2ca}{b^2+2ca}}+\sqrt{1-\frac{2ab}{c^2+2ab}}\)
\(\le\sqrt{1-\frac{2bc}{a^2+b^2+c^2}}+\sqrt{1-\frac{2ca}{a^2+b^2+c^2}}\)\(+\sqrt{1-\frac{2ab}{a^2+b^2+c^2}}\).
\(=\frac{\sqrt{a^2+\left(b-c\right)^2}+\sqrt{b^2+\left(c-a\right)^2}+\sqrt{c^2+\left(a-b\right)^2}}{\sqrt{a^2+b^2+c^2}}\)\(\le\frac{\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}}{\sqrt{a^2+b^2+c^2}}=\frac{a+b+c}{\sqrt{a^2+b^2+c^2}}\)\(\le\frac{a+b+c}{\sqrt{ab+bc+ca}}\).
Dấu \("="\)xảy ra \(\Leftrightarrow a=b=c>0\).
Vậy với \(a,b,c>0\)thì :
\(\frac{a}{\sqrt{a^2+2bc}}+\frac{b}{\sqrt{b^2+2ca}}+\frac{c}{\sqrt{c^2+2ab}}\le\frac{a+b+c}{\sqrt{ab+bc+ca}}\).