đánh giá đại diện..:$\frac{1}{2(a+b-1)+c+d}=\frac{1}{(a+b)+(c+d)+(b+c)-2}\leq \frac{1}{2}.\frac{1}{\sqrt{ab}+\sqrt{cd}+\sqrt{bc}-1}$$\leq \frac{1}{2}.\frac{1}{2\sqrt[4]{abcd}+\sqrt{bc}-1}=\frac{1}{2}.\frac{1}{\sqrt{bc}+1}$tương tự, cộng lại ta đc:$VT\leq \frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{1}{\sqrt{cd}+1}+\frac{1}{\sqrt{bc}+1}+\frac{1}{\sqrt{da}+1})$ $=\frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{\sqrt{ab}}{\sqrt{ab}+1}+\frac{1}{\sqrt{cd}+1}+\frac{\sqrt{cd}}{\sqrt{cd}+1})$ $=1$suy ra dpcm
đánh giá đại diện..:$\frac{1}{2(b
+c-1)+
a+d}=\frac{1}{(a+b)+(c+d)+(b+c)-2}\leq \frac{1}{2}.\frac{1}{\sqrt{ab}+\sqrt{cd}+\sqrt{bc}-1}$$\leq \frac{1}{2}.\frac{1}{2\sqrt[4]{abcd}+\sqrt{bc}-1}=\frac{1}{2}.\frac{1}{\sqrt{bc}+1}$tương tự, cộng lại ta đc:$VT\leq \frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{1}{\sqrt{cd}+1}+\frac{1}{\sqrt{bc}+1}+\frac{1}{\sqrt{da}+1})$ $=\frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{\sqrt{ab}}{\sqrt{ab}+1}+\frac{1}{\sqrt{cd}+1}+\frac{\sqrt{cd}}{\sqrt{cd}+1})$ $=1$suy ra dpcm