Menu Close

if-x-y-z-gt-0-x-y-z-1-and-1-6-then-y-z-x-3-yz-6-1-




Question Number 151215 by mathdanisur last updated on 19/Aug/21
if  x;y;z>0 ; x+y+z=1 and λ≥(1/6) then:  𝛌 Σ ((y + z)/x) + 3 Σ yz ≥ 6𝛌 + 1
$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:;\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}\:\mathrm{and}\:\lambda\geqslant\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{then}: \\ $$$$\boldsymbol{\lambda}\:\Sigma\:\frac{\mathrm{y}\:+\:\mathrm{z}}{\mathrm{x}}\:+\:\mathrm{3}\:\Sigma\:\mathrm{yz}\:\geqslant\:\mathrm{6}\boldsymbol{\lambda}\:+\:\mathrm{1} \\ $$
Answered by dumitrel last updated on 19/Aug/21
p=1  p^2 ≥3q⇒q≤(1/3)≤2λ  q^2 ≥3pr⇒(q/r)≥(3/q)  λΣ(((y+z)/x)+1−1)+3q≥6λ+1⇔λΣ((x+y+z)/x)−3λ+3q≥6λ+1⇔  λ(q/r)+3q≥9λ+1  λ(q/r)+3q≥λ(3/q)+3q  prove ((3λ)/q)+3q≥9λ+1⇔(3q−1)(q−3λ)≥0 true
$${p}=\mathrm{1} \\ $$$${p}^{\mathrm{2}} \geqslant\mathrm{3}{q}\Rightarrow{q}\leqslant\frac{\mathrm{1}}{\mathrm{3}}\leqslant\mathrm{2}\lambda \\ $$$${q}^{\mathrm{2}} \geqslant\mathrm{3}{pr}\Rightarrow\frac{{q}}{{r}}\geqslant\frac{\mathrm{3}}{{q}} \\ $$$$\lambda\Sigma\left(\frac{{y}+{z}}{{x}}+\mathrm{1}−\mathrm{1}\right)+\mathrm{3}{q}\geqslant\mathrm{6}\lambda+\mathrm{1}\Leftrightarrow\lambda\Sigma\frac{{x}+{y}+{z}}{{x}}−\mathrm{3}\lambda+\mathrm{3}{q}\geqslant\mathrm{6}\lambda+\mathrm{1}\Leftrightarrow \\ $$$$\lambda\frac{{q}}{{r}}+\mathrm{3}{q}\geqslant\mathrm{9}\lambda+\mathrm{1} \\ $$$$\lambda\frac{{q}}{{r}}+\mathrm{3}{q}\geqslant\lambda\frac{\mathrm{3}}{{q}}+\mathrm{3}{q} \\ $$$${prove}\:\frac{\mathrm{3}\lambda}{{q}}+\mathrm{3}{q}\geqslant\mathrm{9}\lambda+\mathrm{1}\Leftrightarrow\left(\mathrm{3}{q}−\mathrm{1}\right)\left({q}−\mathrm{3}\lambda\right)\geqslant\mathrm{0}\:{true} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 19/Aug/21
Nice Ser, Thank You
$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *