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Question Number 128005 by mnjuly1970 last updated on 03/Jan/21

$$....nicecalculus...$$ $$ifb>a>0thenprove::$$ $$\frac{a+b}{2}>\frac{b-a}{ln\left(b\right)-ln\left(a\right)}>\sqrt{ab}$$ $$............................$$

Answered by mindispower last updated on 06/Jan/21

$$\iff \frac{\frac{b}{a}+1}{2}>\frac{\frac{b}{a}-1}{ln\left(\frac{b}{a}\right)}>\sqrt{\frac{b}{a}}$$ $$\frac{b}{a}=x,x>1$$ $$\Rightarrow \frac{x+1}{2}=\frac{x-1}{ln\left(x\right)}>\sqrt{x}$$ $$\iff ln\left(x\right)>\frac{2(x-1)}{x+1}..1$$ $$\mathrm{\&}ln\left(x\right)<\sqrt{x}-\frac{1}{\sqrt{x}},..2$$ $$1\iff f\left(x\right)=ln\left(x\right)-2+\frac{4}{x+1}>0$$ $$f^{\prime }\left(x\right)=\frac{1}{x}-\frac{4}{{(x+1)}^2}=\frac{{(x-1)}^2}{x{(x+1)}^2}>0$$ $$f\left(x\right)>f\left(1\right),\mathrm{\forall }x>1$$ $$f\left(1\right)=ln\left(1\right)-2+\frac{4}{2}=0$$ $$\Rightarrow 1..true$$ $$2\iff g\left(x\right),\sqrt{x}-\frac{1}{\sqrt{x}}-ln\left(x\right)>0$$ $$g^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{1}{2x\sqrt{x}}-\frac{1}{x}=\frac{x+1-2\sqrt{x}}{2x\sqrt{x}}=\frac{{(\sqrt{x}-1)}^2}{2x\sqrt{x}}$$ $$g^{\prime }\left(x\right)>0,g\left(1\right)=0\Rightarrow g\left(x\right)>g\left(1\right)=0,\mathrm{\forall }x>1$$ $$2..true$$ $$\Rightarrow \frac{b-a}{2}>\frac{b-a}{ln\left(b\right)-ln\left(a\right)}>\sqrt{ab},\mathrm{\forall }a,bb>a>0$$ $$$$ $$$$

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