$\sqrt{2-x}+\sqrt{2+2x}$ max at 3 when x =1 $\sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1}$ min at 3 when x=1 so $3 \leq \sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1}= \sqrt{2-x}+\sqrt{2+2x} \leq 3$ meaning only $x=1$

StarLex1 wrote: $\sqrt{2-x}+\sqrt{2+2x}$ max at 3 when x =1 $\sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1}$ min at 3 when x=1 so $3 \leq \sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1}= \sqrt{2-x}+\sqrt{2+2x} \leq 3$ meaning only $x=1$ Note that a full proof is required for the exam, but I've forgotten where I've put mine...

hmm i think it is quite clear since $(\sqrt{2-x}+\sqrt{2+2x})^2 \leq (1+2)(2-x+1+x)$ where the equality occur when $\frac{\sqrt{1+x}}{\sqrt{2-x}} = \frac{\sqrt{2}}{1}$ and for the $\sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1} \geq \sqrt{\frac{x^2+1}{2}}+1+\frac{2}{x+1}\geq \frac{x+1}{2}+\frac{2}{x+1}+1 \geq 3$

StarLex1 wrote: hmm i think it is quite clear since $(\sqrt{2-x}+\sqrt{2+2x})^2 \leq (1+2)(2-x+1+x)$ where the equality occur when $\frac{\sqrt{1+x}}{\sqrt{2-x}} = \frac{\sqrt{2}}{1}$ and for the $\sqrt{\frac{x^4+1}{x^2+1}}+\frac{x+3}{x+1} \geq \sqrt{\frac{x^2+1}{2}}+1+\frac{2}{x+1}\geq \frac{x+1}{2}+\frac{2}{x+1}+1 \geq 3$ Yeah I understand this part (I wrote a 2 page proof for the actual comp...), then realized there were easier ways to go about doing this...

Let $x \in(-1, 2]. $ Prove that$$ \frac{x + 3}{x + 1}+\sqrt{\frac{x^4 + 1}{x^2 + 1}}+\sqrt{2 - x}-\sqrt{2 + 2x} \geq\frac{5}{3}+\sqrt{\frac{17}{5}}-\sqrt{6} $$$$ \frac{x + 3}{x + 1}-\sqrt{\frac{x^4 + 1}{x^2 + 1}}+\sqrt{2 - x}+\sqrt{2 + 2x} \geq\frac{5}{3}-\sqrt{\frac{17}{5}}+\sqrt{6} $$