<s>[INST] Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$ 
Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$

$\bf{My\; Try::}$ For $\min$ of $f(x)$
$$\left(\sqrt{13-x}+\sqrt{x}\right)^2=13-x+x+2\sqrt{x}\sqrt{13-x}= 13+2\sqrt{x}\sqrt{13-x}\geq 13$$
Now $$\sqrt{x+27} + \sqrt{13-x}+\sqrt{x} \geq \sqrt{27} + \sqrt{13}$$     
and equality hold at $x=0$
Now How can i calculate $\max$ of $f(x)\;,$ Help required, Thanks
 [/INST] Hint:by Cauchy-Schwarz inequality  $$121=[(x+27)+3(13-x)+2x][1+\dfrac{1}{3}+\dfrac{1}{2}]\ge f^2(x)$$
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