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Question Number 149870 by mathdanisur last updated on 07/Aug/21

if x;y;z;m;n∈R^+ then: Σ_(cyc) (b^(−1) /((m(√x) + n(√y))^2 )) ≥ (3/((m + n)^2 ))

$$\mathrm{if}\:\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{m};\mathrm{n}\in\mathbb{R}^{+} \:\:\mathrm{then}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{b}^{−\mathrm{1}} }{\left(\mathrm{m}\sqrt{\mathrm{x}}\:+\:\mathrm{n}\sqrt{\mathrm{y}}\right)^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\left(\mathrm{m}\:+\:\mathrm{n}\right)^{\mathrm{2}} } \\ $$

Commented by mathdanisur last updated on 08/Aug/21

Sorry dear Ser... Σ_(cyc) (y^(−1) /((m(√x) + n(√y))^2 )) ≥ (3/((m + n)^2 ))

$$\mathrm{Sorry}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}... \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\boldsymbol{\mathrm{y}}^{−\mathrm{1}} }{\left(\boldsymbol{\mathrm{m}}\sqrt{\boldsymbol{\mathrm{x}}}\:+\:\boldsymbol{\mathrm{n}}\sqrt{\boldsymbol{\mathrm{y}}}\right)^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\left(\boldsymbol{\mathrm{m}}\:+\:\boldsymbol{\mathrm{n}}\right)^{\mathrm{2}} } \\ $$

Commented by mathdanisur last updated on 08/Aug/21

Σ_(cyc) (y^(−1) /((m(√x)+n(√y))^2 ))≥^(am-gm) Σ_(cyc) (1/((((m(x+y))/2)+ny)^2 )) = 4Σ_(cyc) (1/((mx+(m+2n)y)^2 )) ≥ 4∙(((1+1+1)^3 )/((Σ_(cyc) (mx+(m+2n)y)^2 )) = ((27)/((m+n)^2 (x+y+z)^2 )) (1) 3=x^2 +y^2 +z^2 ≥(((x+y+z)^2 )/3) ⇔ 9≥(x+y+z)^2 (2) (1) and (2) we obtain Σ_(cyc) (y^(−1) /((m(√x)+n(√y))^2 ))≥((27)/((n+m)^2 ∙9))=(3/((m+n)^2 )) Dear Ser, wrong or true.?

$$\underset{{cyc}} {\sum}\frac{{y}^{−\mathrm{1}} }{\left({m}\sqrt{{x}}+{n}\sqrt{{y}}\right)^{\mathrm{2}} }\overset{{am}-{gm}} {\geqslant}\underset{{cyc}} {\sum}\frac{\mathrm{1}}{\left(\frac{{m}\left({x}+{y}\right)}{\mathrm{2}}+{ny}\right)^{\mathrm{2}} }\:=\:\mathrm{4}\underset{{cyc}} {\sum}\frac{\mathrm{1}}{\left({mx}+\left({m}+\mathrm{2}{n}\right){y}\right)^{\mathrm{2}} } \\ $$$$\geqslant\:\mathrm{4}\centerdot\frac{\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} }{\left(\underset{{cyc}} {\sum}\left({mx}+\left({m}+\mathrm{2}{n}\right){y}\right)^{\mathrm{2}} \right.}\:=\:\frac{\mathrm{27}}{\left({m}+{n}\right)^{\mathrm{2}} \left({x}+{y}+{z}\right)^{\mathrm{2}} }\:\left(\mathrm{1}\right) \\ $$$$\mathrm{3}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}}\:\Leftrightarrow\:\mathrm{9}\geqslant\left({x}+{y}+{z}\right)^{\mathrm{2}} \:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{we}\:{obtain}\:\underset{{cyc}} {\sum}\frac{{y}^{−\mathrm{1}} }{\left({m}\sqrt{{x}}+{n}\sqrt{{y}}\right)^{\mathrm{2}} }\geqslant\frac{\mathrm{27}}{\left({n}+{m}\right)^{\mathrm{2}} \centerdot\mathrm{9}}=\frac{\mathrm{3}}{\left({m}+{n}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Dear}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{wrong}\:\mathrm{or}\:\mathrm{true}.? \\ $$

Commented by mathdanisur last updated on 08/Aug/21

Thank you very much dear Ser...

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\boldsymbol{\mathrm{Ser}}... \\ $$

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