if-x-y-z-gt-0-and-1-1-x-1-1-y-1-1-z-1-then-prove-that-x-y-z-3-4-xyz-

Question Number 161280 by HongKing last updated on 15/Dec/21

if x;y;z>0 and (1/(1+x)) + (1/(1+y)) + (1/(1+z)) = 1 then prove that: x + y + z ≥ (3/4) xyz

$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}}\:=\:\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{xyz} \\ $$

Answered by 1549442205PVT last updated on 16/Dec/21

From the hypothesis (1/(1+x)) + (1/(1+y)) + (1/(1+z)) = 1 we get xy+yz+zx+2(x+y+z)+3=xyz+xy+yz+zx+x+y+z+1 or x+y+z+2=xyz.Hence,the inequality which we need prove to be equivalent to 4(x+y+z)≥3(x+y+z+2)⇔x+y+z≥6(∗) From the hypothesis x+y+z+2=xyz, applying the inequality AM−GM we have x+y+z+2=xyz≤(((x+y+z)/3))^3 ⇒a^3 −27a−54≥0⇒(a−6)(a^2 +6a+9)≥0 ⇔(a−6)(a+3)^2 ≥0⇒a=x+y+z≥6 Thus,the inequality (∗)proved,so x+y+z≥(3/4)xyz(q.e.d) The equality occurs if and only if x=y=z=2

$${From}\:{the}\:{hypothesis}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}}\:=\:\mathrm{1} \\ $$$${we}\:{get}\:{xy}+{yz}+{zx}+\mathrm{2}\left({x}+{y}+{z}\right)+\mathrm{3}={xyz}+{xy}+{yz}+{zx}+{x}+{y}+{z}+\mathrm{1} \\ $$$${or}\:{x}+{y}+{z}+\mathrm{2}={xyz}.{Hence},{the}\:{inequality} \\ $$$${which}\:{we}\:{need}\:{prove}\:{to}\:{be}\:{equivalent}\:{to}\: \\ $$$$\mathrm{4}\left({x}+{y}+{z}\right)\geqslant\mathrm{3}\left({x}+{y}+{z}+\mathrm{2}\right)\Leftrightarrow{x}+{y}+{z}\geqslant\mathrm{6}\left(\ast\right) \\ $$$${From}\:{the}\:{hypothesis}\:{x}+{y}+{z}+\mathrm{2}={xyz}, \\ $$$${applying}\:{the}\:{inequality}\:{AM}−{GM}\:{we} \\ $$$${have}\:{x}+{y}+{z}+\mathrm{2}={xyz}\leqslant\left(\frac{{x}+{y}+{z}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{3}} −\mathrm{27}{a}−\mathrm{54}\geqslant\mathrm{0}\Rightarrow\left({a}−\mathrm{6}\right)\left({a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{9}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\left({a}−\mathrm{6}\right)\left({a}+\mathrm{3}\right)^{\mathrm{2}} \geqslant\mathrm{0}\Rightarrow{a}={x}+{y}+{z}\geqslant\mathrm{6} \\ $$$${Thus},{the}\:{inequality}\:\left(\ast\right){proved},{so} \\ $$$${x}+{y}+{z}\geqslant\frac{\mathrm{3}}{\mathrm{4}}{xyz}\left({q}.{e}.{d}\right) \\ $$$${The}\:{equality}\:{occurs}\:{if}\:{and}\:{only}\:{if} \\ $$$${x}={y}={z}=\mathrm{2} \\ $$

Commented by HongKing last updated on 17/Dec/21

$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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