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Inequality 55 (Unknown Author)


If x,y are positive real numbers then prove that

\displaystyle x^y+y^x>1.


The Inequality is obvious for x,y\geq 1. So, let’s consider the case where x,y\in (0,1). We will now divide the Inequality into two cases.

  • Case 1st: x^{y-1}>y^{x-1}. Then it holds that

\displaystyle (x^y+y^x)^{1/x}=\left[y^x\left(\frac{x^y}{y^x}+1\right)\right]^{1/x}=y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}

therefore, according to Bernulli’s Inequality it will hold that

\displaystyle \begin{aligned}y\cdot\left[\frac{x^y}{y^x}+1\right]^{1/x}&\geq y\left(1+\frac{x^{y-1}}{y^x}\right)\\&\geq y\left(1+\frac{y^{x-1}}{y^x}\right)\\&=y+1>1.\end{aligned}

  • Case 2nd: y^{x-1}>x^{y-1}. Then it is

\displaystyle (x^y+y^x)^{1/y}=\left[x^y\left(\frac{y^x}{x^y}+1\right)\right]^{1/y}=x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}

therefore, once again from Bernulli’s Inequality we get

\displaystyle \begin{aligned}x\cdot\left[\frac{y^x}{x^y}+1\right]^{1/y}&\geq x\left(1+\frac{y^{x-1}}{x^y}\right)\\&\geq x\left(1+\frac{x^{y-1}}{x^y}\right)\\&=x+1>1\end{aligned}

and the proof is completed Q.E.D.


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