holobleep

Có:$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}=\frac{1}{2}(\frac{a}{2-a}+1+\frac{b}{2-b}+1+\frac{c}{2-c}+1)=\frac{1}{2}(\frac{a^{2}}{2a-a^{2}}+\frac{b^{2}}{2b-b^{2}}+\frac{c^{2}}{2c-c^{2}})+\frac{3}{2}\geqslant \frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}$

Ta sẽ chứng minh:$\frac{1}{2}.\frac{(a+b+c)^{2}}{2(a+b+c)-a^{2}-b^{2}-c^{2}}+\frac{3}{2}\geqslant 3\Leftrightarrow (a+b+c)^{2}+9 \geqslant 6(a+b+c)\Leftrightarrow (a+b+c-3)^{2}\geqslant 0$( luôn đúng) suy ra đpcm

Dấu bằng xảy ra khi a=b=c=1