Как доказать лемму a/(b+c)+b/(a+c)+c/(a+b) больше = 3/2

[latex]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}[/latex] [latex]\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1 \geq \frac{3}{2}+3[/latex] [latex]\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b} \geq \frac{9}{2}[/latex] [latex](a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}) \geq \frac{9}{2}[/latex] [latex](\frac{a+b}{2}+\frac{a+c}{2}+\frac{b+c}{2})(\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{a+c}) \geq \\\\3*\sqrt[3] {\frac{a+b}{2}*\frac{a+b}{2}*\frac{b+c}{2}}*3\sqrt[3] {\frac{2}{a+b}*\frac{2}{b+c}*\frac{2}{a+c}}=\\\\9 \geq \frac{9}{2}[/latex]