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$\Leftrightarrow 2 cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\leq \frac{3}{2}\Leftrightarrow 4sin^2\frac{C}{2}-4sin\frac{C}{2}cos\frac{A-B}{2}+1\geq 0\Leftrightarrow (2sin\frac{C}{2}-cos\frac{A-B}{2})^2+sin^2\frac{A-B}{2}\geq 0\rightarrow dpcm.\:Dau \:bang \:xay \: ra\: \Leftrightarrow \Delta ABC\: deu$

VT = $cosA+cosB-cos(A+B)=cosA+cosB-cosA.cosB+sinA.sinB=cosB+[(1-cosB)cosA+sinA.sinB]$

Ta có $a.sinx+b.cosx\leq \sqrt[]{a^2+b^2}$

Vậy VT $\leq \sqrt[]{(1-cosB)^2+sin^2B}=cosB+\sqrt{2-2cosB}=2sin\frac{B}{2}+1-2sin^2\frac{B}{2}=\frac{3}{2}-2(\frac{1}{2}-sin\frac{B}{2})^2\leq \frac{3}{2}$

Đẳng thức xảy ra khi và chỉ khi:

$\left\{\begin{matrix} sin\frac{B}{2} & = &\frac{1}{2} \\ (1-cosB)cosA+sinA.sinB& = & 2sin\frac{B}{2} \end{matrix}\right.$

$\Leftrightarrow$$\left\{\begin{matrix} B & = &\frac{\Pi }{3} \\ \frac{1}{2}cosA+\frac{\sqrt{3}}{2}sinA& = & 2.\frac{1}{2} \end{matrix}\right.$

$\Leftrightarrow A= B= \frac{\Pi }{3}$

$\Leftrightarrow\Delta ABC$ đều