（x+y+z）^2≥3(xy+yz+zx)

Let me expand on chinacai's post... It is easy to see $S=\frac{1}{2}ad_a+\frac{1}{2}bd_b+\frac{1}{2}cd_c=\frac{1}{2}(ad_a+bd_b+cd_c)$. Hence $ \frac{4S^{2}}{3} \ge abd_{a}d_{b}+bcd_{b}d_{c}+cad_{c}d_{a}$ $\iff \frac{4((\frac{1}{2})(ad_a+bd_b+cd_c))^2}{3} \ge abd_{a}d_{b}+bcd_{b}d_{c}+cad_{c}d_{a}$ $\iff \frac{a^2d_a^2+b^2d_b^2+c^2d_c^2+2abd_ad_b+2bcd_bd_c+2cad_cd_a}{3} \ge a$$bd_{a}d_{b}+bcd_{b}d_{c}+cad_{c}d_{a}$ $\iff a^2d_a^2+b^2d_b^2+c^2d_c^2\ge abd_{a}d_{b}+bcd_{b}d_{c}+cad_{c}d_{a}$ Which is equivalent to $x^2+y^2+z^2\ge xy+yz+zx$. Equality holds, and hence the $LHS$ attains it's maximum, only when $x=y=z$ which occurs in the case $M=G$. IMOs have gotten harder!