Solve the inequality

Let sin x = t

⇒ cos 2x = 1 – 2t²

⇒ the inequality is : t + 1 – 2t² > 1

⇒ t – 2t² > 0

⇒ 2t² – t < 0

⇒ t(2t – 1) < 0

⇒ (t – 0) (t – 1/2) < 0

⇒ 0 < t < 1/2
⇒ 0 < sin x < 1/2

In 0 ≤ x ≤ π/2, this means that 0 < x < π/6 is the solution

$\textbf{Here $\mathbf{A+B+C=\pi}$ and $\mathbf{0\leq A,B,C\leq \mathbf{\pi}}}$\\\\ Now Here Given that $\mathbf{cos\;A+\underbrace{\mathbf{cos\:B+cos\:C}}=\frac{3}{2}}$\\\\\\ $\mathbf{1-2sin^2\left(\frac{A}{2}\right)+2\cdot cos\left(\frac{B+C}{2}\right)\cdot cos\left(\frac{B-C}{2}\right)=\frac{3}{2}}$\\\\\\ $\mathbf{1-2\cdot sin^2\left(\frac{A}{2}\right)+2.sin\left(\frac{A}{2}\right).cos\left(\frac{B-C}{2}\right)=\frac{3}{2}}$\\\\\\ $\mathbf{4.sin^2\left(\frac{A}{2}\right)-4.sin\left(\frac{A}{2}\right).cos\left(\frac{B-C}{2}\right)+1=0}$\\\\\\ $\textbf{For Real Roots} \mathbf{Here\ D\geq 0}$\\\\ \textbf{So $\mathbf{16.cos^2\left(\frac{B-C}{2}\right)-4.4.1\geq 0}$}\\\\\\ So $\mathbf{cos^2\left(\frac{B-C}{2}\right)\geq 1\Leftrightarrow cos^2\left(\frac{B-C}{2}\right)=1\Leftrightarrow cos\left(\frac{B-C}{2}\right)=1}$\\\\\\ $\mathbf{\frac{B-C}{2}=0\Leftrightarrow B=C}$\\\\\\ Similarly We Get $\mathbf{A=C}$\\\\\\ So $\boxed{\boxed{\mathbf{A=B=C=\frac{\pi}{3}}}}$

yes Shubhodip you are saying Right. Using Jesan Inequality We can solve it with in 2 or 3 lines.

Another method

2)

Thanks Omkar for Nice alternative.

but what i said was not wrong either :P

i have recalled how to do it by jensens

note that cosA + cosB ≤ 2 sinC2

so it follows that cosA + cosB + cosC ≤sinA2 + sinB2 + sinC2

By jensens sinA2 + sinB2 + sinC2 ≤32

probably i read that in Arihant's MO book