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How to prove which is bigger, $(x^2+1)^5$ or $(x^5+1)^2$ when $x>0$ ?

Perhaps this is easy question. Thank you.

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closed as off-topic by Grigory M, Thursday, Old John, Hawk, RecklessReckoner Jul 10 at 22:39

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Frankly, since the question assumes that on the range (0, inf) ONE of these is actually bigger, then simply plugging in the number 1 to both of them and seeing which is bigger there should give you the answer. (It's impossible for it bigger for the case of x=1, and yet smaller for all cases of x>0) –  Cruncher Jul 10 at 13:15

9 Answers 9

up vote 14 down vote accepted

Here is another method, not necessarily the most intuitive, but quite elementary to verify: We have that $$ \begin{array}{rcl}(x^2+1)^5-(x^5+1)^2&=&5 x^8+10 x^6-2 x^5+10 x^4+5 x^2\\ &=&x^2((5x^6-2x^3+1/5)+(10x^4+10x^2+(5-1/5)))\\ &=&5x^2(x^3-1/5)^2+10x^2(x^4+x^2+0.48)\\ &\ge&0,\end{array} $$ with equality iff $x=0$.

The point here is that any inequality between polynomials can be proved by verifying that their difference is a sum of squares of rational functions. See here for some references and other examples.

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Another method is to consider the function: $$f(h)=(x^h+1)^{1/h} \Rightarrow \ln f(h)=\frac{1}{h}\ln(x^h+1)$$ Differentiate both the sides with respect to $h$ i.e $$\begin{aligned} \frac{f'(h)}{f(h)} &= \frac{1}{h^2}\left(\frac{h\,x^h\ln x}{x^h+1}-\ln(1+x^h)\right)\\ &=\frac{1}{h^2(x^h+1)}\left(h\,x^h\ln x-(x^h+1)\ln(1+x^h)\right)\\ &=\frac{1}{h^2(x^h+1)}\left(x^h\ln\left(\frac{x^h}{x^h+1}\right)-\ln(1+x^h)\right)\\ &=-\frac{1}{h^2(x^h+1)}\left(x^h\ln\left(1+\frac{1}{x^h}\right)+\ln(1+x^h)\right)\\ \Rightarrow f'(h) &=-\frac{f(h)}{h^2(x^h+1)}\left(x^h\ln\left(1+\frac{1}{x^h}\right)+\ln(1+x^h)\right) \end{aligned}$$

Notice that $f(h)$, $\ln\left(1+\dfrac{1}{x^h}\right)$, $\ln(1+x^h)$ and the denominator are positive so we can conclude that $f'(h)<0$. Hence, $$(x^2+1)^{1/2} >(x^5+1)^{1/5} \Rightarrow \boxed{(x^2+1)^5 > (x^5+1)^2}$$

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Hint: Take the derivative of both functions, and try to evaluate which of these is larger on the interval. This should be very intuitive once you see the derivatives.

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The degrees are the same. –  Jonas Meyer Jul 10 at 4:43
Aryaman Bansal has done the calculation, and I'd hardly call it intuitive. (Ok, so his calculation is wrong, but the correct values aren't any better in this respect.) dmk got the correct values but didn't take the reasoning all the way. The derivatives are rather harder to compare than the originals. –  Gilles Jul 10 at 12:45
@Gilles - What do you mean he has "already done the calculation"? I was the first one who posted an answer to this question, and mine is a hint. I don't see what your point is. –  Aleksander Jul 10 at 19:12
@Aleksander Your post is a hint, it barely qualifies as an answer. And as a hint, it's a bad one: while the technique can work, the statement that “this should be very intuitive once you see the derivatives” is wrong. –  Gilles Jul 10 at 19:25
@Gilles - I will take your comments into consideration. However, providing hints as opposed to a silver platter answer, is actually encouraged. Answers are meant for discussing and/or elaborating on the question/answer. meta.math.stackexchange.com/questions/10589/… –  Aleksander Jul 10 at 19:33

Let us consider the function $$f(x)=\frac{(x^2+1)^5}{(x^5+1)^2}$$ where $x \geq 0$. For $x=0$, $f(1)=0$. The derivative is, after a few simplifications, $$f'(x)=-\frac{10 x \left(x^2+1\right)^4 \left(x^3-1\right)}{\left(x^5+1\right)^3}$$ which only cancels for $x=1$ and $f(1)=8$; at this point, $f''(1)=-60$, so it is a maximum.

Now, look at Taylor expansions around $x=0$; we find $$f(x)=1+5 x^2+O\left(x^3\right) \gt 1$$ For large values of $x$, Taylor expansion is $$f(x)=1+\frac{5}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right) \gt 1$$

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  • If $x\ge1$ then ${\rm LHS}>x^{10}+5x^8+1>{\rm RHS}$.
  • If $0<x<1$ then ${\rm LHS}>x^{10}+5x^2+1>{\rm RHS}$.
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Hint: Look at each function's deriviative. For $\left(x^2 + 1 \right)^5$, it's $10x(x^2+1)^4$ and for $\left( x^5 + 1\right)^2$, it's $10x\cdot x^3(x^5 + 1)$. If $x < 1$, note that the derivative of the first function contains an $x^2$ term, whereas that of the second function doesn't.

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This is not enough, it only shows that $(x^2+1)^5$ is eventually larger, but does not imply the inequality holds for all $x>0$. –  Andres Caicedo Jul 10 at 4:42
Good point. I was starting to look at the derivative cos I suspected that might help. I'll update momentarily. –  dmk Jul 10 at 4:46

Set $x=e^{t/10}$. Then: $$ (1+x^2)^5 = (1+e^{t/5})^5 = e^{t/2}(2\cosh(t/10))^5$$ $$(1+x^5)^2 = (1+e^{t/2})^2 = e^{t/2}(2\cosh(t/4))^4$$ so the inequality $(1+x^2)^5\geq (1+x^5)^2$ follows from the log-convexity of the hyperbolic cosine function, implied by the inequality $\frac{d^2}{dx^2}\log\cosh x\geq 0$: $$\frac{d}{dx}\log\cosh x = \tanh x,\qquad \frac{d}{dx}\tanh x=\frac{1}{\cosh^2 x}\geq 0.$$

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Let the functions be $f(x)$ and $g(x)$ respectively. By differentiating both the functions, we get $10x(x^2 +1)^4$ and $10x^4(x^5 +1)$ respectively. Both are positive for all $x> 0$. This means both are strictly increasing functions.

Now, to compare these differentials, we can compare $(x^2 +1)^4$ and $x^3(x^5 +1)$. By differentiating, we get $8x(x^2 +1)^3$ and $8x^7 + 3x^2$. For $f'(x)$ to be greater than $g'(x)$, it is necessary that $8x^7 + 24x^5 + 24x^3 + 8x> 8x^7 + 3x^2$ => $24x^4 + 24x^2 - 3x + 8> 0$ The minima of the above equation is around 8 which is greater than 0 which implies that the above equation is true for all $x> 0$

This implies that $f'(x)$ is greater than $g'(x)$ for all $x> 0$. Hence, $f(x)$ is greater than $g(x)$ for all $x> 0$.

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The former,


$$(x^5 + 1)^2 = x^{10} + 2x^5 + 1$$


$$(x^2 +1)^5 = x^{10} + 5x^8... +1$$

Subtract the top from the bottom:

$$= 5x^8 + ... + 5x^2 - 2x^5$$

Clearly this is positive for $x$ large enough.

Now proving that greater than $0$ is/isn't the critical number requires an analysis of how quickly each expression grows (since the both have the same exact value at 0).

So we can differentiate both expressions:

$$10x^4(x^5 +1)$$


$$10x(x^2 +1)^4$$

Again clearly the second expression based on something similar to our collecting terms argument above

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I should add, that this won't be sufficient now looking back at it until one performs more derivatives until both are reduced to a linear term plus a constant term –  frogeyedpeas Jul 10 at 4:46

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