It is obvious that if $x \geq 3 $ the LHS will always be bigger. But how can I prove this by induction. I have tried using the binomial theorem for $(x+1)^x$ but I couldn't complete it. The result is always bigger than $2x^2$ but I can't determine an upper bound.
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Is it $x^{(x+1)}\gt (x+1)^x $ ? – A learner Aug 08 '21 at 12:53
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Yes, it is. I didn't write it properly. – İsmim Yok Aug 08 '21 at 12:54
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Why must you use induction? – Ritam_Dasgupta Aug 08 '21 at 12:56
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1Is $x$ a natural number? – Arctic Char Aug 08 '21 at 12:57
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I just wanted to see if it is possible by induction. – İsmim Yok Aug 08 '21 at 12:58
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It's not true for $x=1$. – John Douma Aug 08 '21 at 12:59
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$$(x+1)^{x+2}=x^{x+1}\cdot\left(1+\frac1x\right)^{x+1}\cdot(x+1)=x^{x+1}\cdot\left(1+\frac1x\right)^x\cdot\frac{(x+1)^2}x$$$$(x+2)^{x+1}=(x+1)^x\cdot\left(1+\frac1{x+1}\right)^x\cdot(x+2)$$
Note that $1+\frac1x>1+\frac1{x+1}$ and $(x+1)^2>x\cdot(x+2)$, so we are done by induction (assuming you prove the base case $x=3$).
Rushabh Mehta
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