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Question Number 79644 by loveineq. last updated on 26/Jan/20

$$f\left(x\right)=\frac{4(4x^2+3)}{4x^2+4x+5}$$$$\mathrm{prove}\mathrm{that}f\left(x\right)⩾2.$$

Commented by john santu last updated on 27/Jan/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)<2$$$$\frac{{16\mathrm{x}}^2+12}{{4\mathrm{x}}^2+4\mathrm{x}+5}<2\Rightarrow {4\mathrm{x}}^2+4\mathrm{x}+5>0\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$$${16\mathrm{x}}^2+12<{8\mathrm{x}}^2+8\mathrm{x}+10$$$${8\mathrm{x}}^2-8\mathrm{x}+2<0\Rightarrow \mathrm{\Delta }=64-4.8.2=0$$$${\mathrm{x}}^2-\mathrm{x}+\frac{1}{4}={(\mathrm{x}-\frac{1}{2})}^2⩾0$$$$\mathrm{contradiction}\mathrm{with}\mathrm{the}\mathrm{initial}$$$$\mathrm{statement}.$$$$\mathrm{so}\mathrm{f}\left(\mathrm{x}\right)⩾2.$$$$$$

Answered by mr W last updated on 27/Jan/20

$$f\left(x\right)=\frac{4(4x^2+3)}{4x^2+4x+5}$$$$=\frac{4(4x^2+4x+5-4x-2)}{4x^2+4x+5}$$$$=4(1-\frac{x+\frac{1}{2}}{x^2+x+\frac{5}{4}})$$$$=4(1-\frac{x+\frac{1}{2}}{{(x+\frac{1}{2})}^2+1})$$$$=4(1-\frac{1}{(x+\frac{1}{2})+\frac{1}{(x+\frac{1}{2})}})$$$$=4(1+\frac{1}{t+\frac{1}{t}})⩾4(1-\frac{1}{2})=2$$$$witht=-x-\frac{1}{2}$$

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