Ta biến đổi 1 tí nhé
\(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\ge4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)
Tới đây dễ dàng áp dụng BĐT \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\)
\(\Leftrightarrow\frac{3}{a+b}\le\frac{3}{4}.\frac{1}{a}+\frac{3}{4}.\frac{1}{b}\left(1\right)\)
\(\Leftrightarrow\frac{2}{b+c}\le\frac{1}{2}.\frac{1}{b}+\frac{1}{2}.\frac{1}{c}\left(2\right)\)
\(\Leftrightarrow\frac{1}{a+c}\le\frac{1}{4}.\frac{1}{a}+\frac{1}{4}.\frac{1}{c}\left(3\right)\)
Cộng vế với vế của (1), (2), (3) suy ra
\(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{3}{4}\cdot\frac{1}{a}+\frac{3}{4}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{b}+\frac{1}{2}\cdot\frac{1}{c}+\frac{1}{4}\cdot\frac{1}{a}+\frac{1}{4}\cdot\frac{1}{c}\)
\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{a}+\frac{5}{4}\cdot\frac{1}{b}+\frac{3}{4}\cdot\frac{1}{b}\)
\(\Leftrightarrow\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{4}{a}+\frac{5}{b}+\frac{3}{c}\right)\)
\(\Leftrightarrow Dpcm\)
