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Question-151351

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Question Number 151351 by mathdanisur last updated on 20/Aug/21
Answered by dumitrel last updated on 20/Aug/21
λ(√((a^2 +b^2 )/2))+((2ab)/(a+b))≥((λ+1)/2)(a+b) if a=b true suppose a<b ;(a/b)=t∈(0;1) ⇔λ(√((t^2 +1)/2))+((2t)/(t+1))≥(((λ+1)(t+1))/2) ⇔λ((((√(2t^2 +2))−t−1)/2))≥((t+1)/2)−((2t)/(t+1))⇔ λ∙(((λ−1)^2 )/( (√(2t^2 +2))+t+1))≥(((t−1)^2 )/(t+1))⇔λ≥(((√(2(t^2 +1)))+t+1)/(t+1)) true (√2)>((√(2(t^2 +1)))/(t+1)) λ≥(√2)+1>((√(2(t^2 +1)))/(t+1))+((t+1)/(t+1))=(((√(2(t^2 +1)))+t+1)/(t+1))
$$\:\lambda\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}+\frac{\mathrm{2}{ab}}{{a}+{b}}\geqslant\frac{\lambda+\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$$${if}\:{a}={b}\:{true} \\ $$$${suppose}\:{a}<{b}\:;\frac{{a}}{{b}}={t}\in\left(\mathrm{0};\mathrm{1}\right) \\ $$$$\Leftrightarrow\lambda\sqrt{\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{2}{t}}{{t}+\mathrm{1}}\geqslant\frac{\left(\lambda+\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow\lambda\left(\frac{\sqrt{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}−{t}−\mathrm{1}}{\mathrm{2}}\right)\geqslant\frac{{t}+\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}{t}}{{t}+\mathrm{1}}\Leftrightarrow \\ $$$$\lambda\centerdot\frac{\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}+{t}+\mathrm{1}}\geqslant\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}+\mathrm{1}}\Leftrightarrow\lambda\geqslant\frac{\sqrt{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+{t}+\mathrm{1}}{{t}+\mathrm{1}}\:{true} \\ $$$$\sqrt{\mathrm{2}}>\frac{\sqrt{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}}{{t}+\mathrm{1}} \\ $$$$\lambda\geqslant\sqrt{\mathrm{2}}+\mathrm{1}>\frac{\sqrt{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}}{{t}+\mathrm{1}}+\frac{{t}+\mathrm{1}}{{t}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+{t}+\mathrm{1}}{{t}+\mathrm{1}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 20/Aug/21
Nice Ser, Thank You
$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

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