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Question Number 80039 by jagoll last updated on 30/Jan/20

prove that (1+x)(1+(1/x))≥4

$${prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{4} \\ $$

Commented by mr W last updated on 30/Jan/20

only for x∈R^+ 1+x≥2(√x) 1+(1/x)≥2(1/(√x)) (1+x)(1+(1/x))≥2(√x)×2(1/(√x))=4 or (1+x)(1+(1/x))=2+x+(1/x)≥2+2=4 for x∈R^− (1+x)(1+(1/x))=2+x+(1/x)=2−(−x+(1/(−x)))≤2−2=0

$${only}\:{for}\:{x}\in{R}^{+} \\ $$$$\mathrm{1}+{x}\geqslant\mathrm{2}\sqrt{{x}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2}\frac{\mathrm{1}}{\sqrt{{x}}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{2}\sqrt{{x}}×\mathrm{2}\frac{\mathrm{1}}{\sqrt{{x}}}=\mathrm{4} \\ $$$${or} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}+{x}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2}+\mathrm{2}=\mathrm{4} \\ $$$$ \\ $$$${for}\:{x}\in{R}^{−} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}+{x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}−\left(−{x}+\frac{\mathrm{1}}{−{x}}\right)\leqslant\mathrm{2}−\mathrm{2}=\mathrm{0} \\ $$

Commented by Tony Lin last updated on 30/Jan/20

[1^2 +((√x))^2 ][1^2 +((1/(√x)))^2 ]≥(1+1)^2 =4

$$\left[\mathrm{1}^{\mathrm{2}} +\left(\sqrt{{x}}\right)^{\mathrm{2}} \right]\left[\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\sqrt{{x}}}\right)^{\mathrm{2}} \right]\geqslant\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$

Commented by MJS last updated on 30/Jan/20

y=(1+x)(1+(1/x)) defined for x∈R\{0} (1+x)(1+(1/x))<4 x+(1/x)<2 (1) x>0 x^2 −2x+1<0 (x−1)^2 <0 wrong ⇒ (1+x)(1+(1/x))≥4 (2) x<0 x^2 −2x+1>0 (x−1)^2 >0 true ∀x<0 ⇒ (1+x)(1+(1/x))<4 { (((1+x)(1+(1/x))≥4; x>0)),(((1+x)(1+(1/x))<4; x<0)) :}

$${y}=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)<\mathrm{4} \\ $$$${x}+\frac{\mathrm{1}}{{x}}<\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\:{x}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}<\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} <\mathrm{0}\:\mathrm{wrong}\:\Rightarrow\:\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}>\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0}\:\mathrm{true}\:\forall{x}<\mathrm{0}\:\Rightarrow\:\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)<\mathrm{4} \\ $$$$\begin{cases}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{4};\:{x}>\mathrm{0}}\\{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)<\mathrm{4};\:{x}<\mathrm{0}}\end{cases} \\ $$

Commented by mr W last updated on 30/Jan/20

typo sir? (1+x)(1+(1/x))<4; x<0 should be: (1+x)(1+(1/x))≤0; x<0

$${typo}\:{sir}? \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)<\mathrm{4};\:{x}<\mathrm{0} \\ $$$${should}\:{be}: \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\leqslant\mathrm{0};\:{x}<\mathrm{0} \\ $$

Commented by MJS last updated on 30/Jan/20

You are right but with my method I only examined if it′s <4.

$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}\:\mathrm{but}\:\mathrm{with}\:\mathrm{my}\:\mathrm{method}\:\mathrm{I}\:\mathrm{only} \\ $$$$\mathrm{examined}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:<\mathrm{4}. \\ $$

Commented by jagoll last updated on 31/Jan/20

thank you

$${thank}\:{you}\: \\ $$

Answered by $@ty@m123 last updated on 31/Jan/20

Let y=(1+x)(1+(1/x)) ⇒y=2+x+(1/x) (dy/dx)=1−(1/x^2 ) For max. or min. (dy/dx)=0 1−(1/x^2 )=0 ⇒x=±1 y∣_(x=1) =4, y∣_(x=−1) =0 (d^2 y/dx^2 )=(1/x^3 ) [(d^2 y/dx^2 )]_(x=1) =1>0 ∴ y is mininmum at x=1 and min.(y)=4

$${Let}\:{y}=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow{y}=\mathrm{2}+{x}+\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${For}\:{max}.\:{or}\:{min}. \\ $$$$\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{x}=\pm\mathrm{1} \\ $$$${y}\mid_{{x}=\mathrm{1}} \:=\mathrm{4}, \\ $$$${y}\mid_{{x}=−\mathrm{1}} =\mathrm{0} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$\left[\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right]_{{x}=\mathrm{1}} =\mathrm{1}>\mathrm{0} \\ $$$$\therefore\:{y}\:{is}\:{mininmum}\:{at}\:{x}=\mathrm{1} \\ $$$${and}\:{min}.\left({y}\right)=\mathrm{4} \\ $$

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