bài 1
ta có$\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{a+c+2b}}+\sqrt{\frac{c}{a+b+2c}}\leq \sqrt{3(\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c})}\leq \sqrt{\frac{3}{4}(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c})}=\sqrt{\frac{9}{4}}=\frac{3}{2}$
bài 2
bđt tương đương $\frac{a^{2}-ab+b^{2}}{ab}+\frac{b^{2}+c^{2}-bc}{bc}+\frac{a^{2}-ac+c^{2}}{ac}\geq 2(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$
có $\frac{a^{2}-ab+b^{2}}{ab}+\frac{b^{2}+c^{2}-bc}{bc}+\frac{a^{2}-ac+c^{2}}{ac}\geq \frac{a^{2}+b^{2}}{2ab}+\frac{b^{2}+c^{2}}{2bc}+\frac{c^{2}+a^{2}}{2ac}=\frac{a}{2b}+\frac{b}{2a}+\frac{b}{2c}+\frac{c}{2b}+\frac{c}{2a}+\frac{a}{2c}$
ta có $2(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\leq \frac{1}{2}(\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b})$ đpcm