jred wrote: For $\theta\in[0, \dfrac{\pi}{2}]$, the following inequality $\sqrt{2}(2a+3)\cos(\theta-\dfrac{\pi}{4})+\dfrac{6}{\sin\theta+\cos\theta}-2\sin2\theta<3a+6$ is always true. Determine the range of $a$. Solution: Let $t=\sin\theta+\cos\theta=\sqrt{2}\sin(\theta+\frac{\pi}{4})\in [1,\sqrt{2}]$, then, $\sin 2\theta=t^2-1.$ The given inequality can be transformed into the following $a(2t-3)<2t^2-3t-\frac 6t+4$. Since $2t-3\leq 2\sqrt{2}-3<0$, we have $a>\frac{2t^2-3t-\frac 6t+4}{2t-3}=t+\frac{2}{t}=f(t), t\in [1,\sqrt{2}]$. $f_{max}=f(1)=3.$ Therefore, $a>3$. We have done!

Let $x=\sin \theta+\cos \theta$ We write the original equation to be $(2a+3)x+\frac{6}{x}-2(x^2-1)<3a+6$, which is equivalent to the expression: $2x^2-2ax-3x-\frac{x}{6}+3a+4>0, 2x(x+\frac{2}{x}-a)-3(x+\frac{2}{x}-a)=(2x-3)(x+\frac{2}{x}-a)$ As $\theta\in [0,\frac{\pi}{2}], 1<x<\sqrt{2}$ which means that $2a-3<0$, we need to guarantee that $x+\frac{2}{x}-a<0$, now we need to find $x+\frac{2}{x}$'s maximum over $[1,\sqrt{2}]$, which is reached when $x=1, a>3$ as desired