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\(P=\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)
\(=\frac{6047-a}{2015+a}+\frac{6048-b}{2016+b}+\frac{6049-c}{2017+c}\)
\(=\frac{8062}{2015+a}+\frac{8064}{2016+b}+\frac{8066}{2017+c}-3\)
\(\ge\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{2015+2016+2017+a+b+c}-3=\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{8064}-3\)
Dấu = xảy ra khi ....
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Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:
\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)
\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)
\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)
Cộng (1),(2) và (3) có:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)