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Question Number 179090 by CrispyXYZ last updated on 24/Oct/22

$$\mathrm{Given}a>0,b>0,c>2,a+b=1.$$$$\mathrm{find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\frac{3ac}{b}+\frac{c}{ab}+\frac{6}{c-2}.$$

Commented by Frix last updated on 24/Oct/22

$$24$$

Answered by MJS_new last updated on 24/Oct/22

$$a>0\wedge b>0\wedge a+b=1\Rightarrow 0<a<1\wedge 0<b<1$$$$\Rightarrow \mathrm{let}a={\sin }^{2}x\wedge b={\cos }^{2}x$$$$\mathrm{I}\mathrm{prefer}x=\arctan t\Rightarrow a=\frac{t^2}{t^2+1}\wedge b=\frac{1}{t^2+1}$$$$c>2\Rightarrow c=2+u^2$$$$$$$$\Rightarrow$$$$$$$$\frac{3ac}{b}+\frac{c}{ab}+\frac{6}{c-2}=$$$$=3t^2(u^2+2)+\frac{{(t^2+1)}^2(u^2+2)}{t^2}+\frac{6}{u^2}=f(t,u)$$$$\frac{df}{dt}=0$$$$6(u^2+2)t+\frac{2(t^4-1)(u^2+2)}{t^3}=0$$$$2(4t^4-1)(u^2+2)=0$$$$t=\pm \frac{\sqrt{2}}{2}$$$$f(\pm \frac{\sqrt{2}}{2},u)=\frac{3(u^2+2)}{2}+\frac{9(u^2+2)}{2}+\frac{6}{u^2}=\frac{6(u^2+1)}{u^2}$$$$\frac{df}{du}=0$$$$\frac{12(u^4-1)}{u^3}=0$$$$u=\pm 1$$$$f(\pm \frac{\sqrt{2}}{2},\pm 1)=24$$$$\mathrm{with}a=\frac{1}{3}\wedge b=\frac{2}{3}\wedge c=3$$

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