ĐK[tex]y\geq z\geq x\geq 0[/tex]
Áp dụng BĐT [tex]AM - GM[/tex] ta có:
[tex]\sqrt{x}\leq \frac{x+1}{2} (1)[/tex]
[tex]\sqrt{y-z}\leq \frac{y-z+1}{2} (2)[/tex]
[tex]\sqrt{z-x}\leq \frac{z-x+1}{2} (3)[/tex]
Từ (1) , (2), (3) suy ra ta có :
[tex]\sqrt{x}+\sqrt{y-z}\sqrt{z-x}\leq \frac{x+1y-z+1+z-x+1}{2} = \frac{1}{2}(y+3)[/tex]
Dấu '' = '' xảy ra khi : y = 3, z=2, x = 1