jasperE3 wrote: Let $a,b,c$ be the real numbers with $a\ge\frac85b>0$ and $a\ge c>0$. Prove the inequality $$\frac45\left(\frac1a+\frac1b\right)+\frac2c\ge\frac{27}2\cdot\frac1{a+b+c}.$$ $a=2, b=1, c=1$ doesn't satisfy the inequality. $$\frac{4}{5}(\frac{1}{a}+\frac{1}{b})+\frac{2}{c}=\frac{4}{5}(\frac{1}{2}+\frac{1}{1})+\frac{2}{1}=\frac{4}{5}(\frac{3}{2})+2=\frac{6}{5}+2=3,2$$$$\frac{27}{2}\cdot\frac{1}{a+b+c}=\frac{27}{2}\cdot\frac{1}{2+1+1}=\frac{27}{2}\cdot\frac{1}{4}=\frac{27}{8}=3,375$$