Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:


share|cite|improve this question
I changed the title from "Evaluating" to "Prove that". "Evaluating" seems inappropriate. Hope it is fine with you. – user17762 Jun 13 '12 at 8:31
up vote 12 down vote accepted

Let $a = 1+x$ and $b = 1+y$. Then we need to prove that $$\dfrac{(x+1)^2}{y} + \dfrac{(y+1)^2}{x} \geq 8$$ i.e. $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$ for $x,y \geq 0$.

Now apply AM-GM as shown below. $$\dfrac{\dfrac{x^2}{y} + \dfrac{x}{y} + \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + \dfrac{y}{x} + \dfrac{y}{x} + \dfrac1x}{8} \geq \sqrt[8]{\dfrac{x^2}{y} \times \dfrac{x}{y} \times \dfrac{x}{y} \times \dfrac1y \times \dfrac{y^2}{x} \times \dfrac{y}{x} \times \dfrac{y}{x} \times \dfrac1x} = 1$$ Hence, we get that $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$ which is what we wanted to show.

share|cite|improve this answer

You have $a^2 \geq 4(a-1)$ and $b^2 \geq 4(b-1)$. Then

$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq 4 \frac{a-1}{b-1}+4 \frac{b-1}{a-1}\geq8 \,,$$

the last inequality being either AM-GM, or the standard $x +\frac{1}{x} \geq 2$.

share|cite|improve this answer


share|cite|improve this answer
Nice solution. A short comment $a^2(a-1)+b^2(b-1) \geq a^2(b-1)+b^2(a-1)$ is equivalent to $a^3+b^3 \geq a^2b+ab^2$ which follows immediately from AM-GM or $(a-b)^2(a+b) \geq 0$. – N. S. Jun 13 '12 at 21:17
(+1) Just one comment : One can reach $\frac{a^2}{b - 1} + \frac{b^2}{a - 1} \geq \frac{a^2}{a - 1} + \frac{b^2}{b - 1}$ directly via Rearrangement Inequality – TenaliRaman Jun 13 '12 at 21:18
Thank you, indeed. I was just trying to use "minimal" known constructions to get to the answer (that is am-gm and reciprocals) – Valentin Jun 13 '12 at 21:21
@Valentin I realized that :-). I just wanted to throw that comment in, instead of writing up a new answer, given that it pretty much would follow your steps anyways. – TenaliRaman Jun 13 '12 at 21:28

Alternatively, you can use calculus and consider f(a,b) to minimize. You will find that happens for a=b=2.

share|cite|improve this answer
Yeah. If you're willing you may post the whole proof. – user 1618033 Jun 13 '12 at 9:06

If you don't mind I give the details of @Vasea Igor's answer. Define $f(a,b):=\frac{a^2}{b-1}+\frac{b^2}{a-1}$, where $a,b>1$. Then $$ \frac{\partial}{\partial a}f(a,b)=\frac{2a(a-1)^2-(b-1)b^2}{(b-1)(a-1)^2}=\frac{2a^3-4a^2+2a-b^3+b^2}{(b-1)(a-1)^2}, $$ $$ \frac{\partial}{\partial b}f(a,b)=-\frac{a^2}{(b-1)^2}+\frac{2b}{a-1}=-\frac{a^3-a^2-2b^3+4b^2-2b}{(b-1)^2(a-1)}. $$ The real solutions of the system $$ 2a^3-4a^2+2a-b^3+b^2=0, $$ $$ a^3-a^2-2b^3+4b^2-2b=0 $$ are $$ (a,b)={(2,2),(0,0),(1,0),(0,1),(1,1)}. $$ Since $a,b>1$ the only candidate is $(a,b)=(2,2)$. Now we compute $H(a,b)$ the Hesse Matrix of $f(a,b)$ at $a=2, b=2$. $$ H(a,b)=\left[ \begin {array}{cc} \frac{2}{b-1}+2\,{\frac {{b} ^{2}}{ \left( a-1 \right) ^{3}}}&-2\,{\frac {a}{ \left( b-1 \right) ^{ 2}}}-2\,{\frac {b}{ \left( a-1 \right) ^{2}}}\\ -2\, {\frac {a}{ \left( b-1 \right) ^{2}}}-2\,{\frac {b}{ \left( a-1 \right) ^{2}}}&2\,{\frac {{a}^{2}}{ \left( b-1 \right) ^{3}}}+\frac{2}{a-1}\end {array} \right] . $$ Now we get $$ H(2,2)=\left[ \begin {array}{cc} 10&-8\\ -8&10\end {array} \right] . $$ The eigenvalues of $H(2,2)$ are $18,2>0$. It means $H(2,2)$ is positive definite, which implies the function $f$ has a local minimum (and at the same time, global minimum) at $a=2, b=2$ which is $f(2,2)=8$.

share|cite|improve this answer

WLOG assume $a \leq b$, therefore, $a^2 \leq b^2$ and $\frac{1}{b - 1} \leq \frac{1}{a - 1}$. Therefore, by Rearrangement Inequality, $$\frac{a^2}{b-1} + \frac{b^2}{a - 1} \geq \frac{a^2}{a-1} + \frac{b^2}{b - 1}$$ From here, one can simply follow Valentin's steps, $$\frac{a^2}{a - 1} + \frac{b^2}{b - 1} = 4 + (a - 1) + \frac{1}{a - 1} + (b - 1) + \frac{1}{b - 1} \geq 4 + 2 + 2 = 8$$


As Chris points out, $(a - 2)^2 \geq 0$ implies $a^2 - 4a + 4 \geq 0$ or $a^2 \geq 4(a - 1)$ and hence $\frac{a^2}{a - 1} \geq 4$. Thereby, cutting short few more steps.

share|cite|improve this answer


$\frac{a^2}{b-1}+4(b-1)\geq 4a$

$\frac{b^2}{a-1}+4(a-1)\geq 4b$

$\Rightarrow \frac{a^2}{b-1}+4(b-1)+\frac{b^2}{a-1}+4(a-1)\geq 4a+4b$

$\Rightarrow \frac{a^2}{b-1}+\frac{b^2}{a-1}\geq 8$

share|cite|improve this answer

As Marvis indicated, the problem is equivalent to showing that for $x, y > 0$ one has $${(1 + x)^2 \over y} + {(1 + y)^2 \over x} \geq 8$$ To make the terms of the right homogeneity, it's natural to use AM-GM in the numerators in the form $1 + x \geq 2\sqrt{x}$ and $1 + y \geq 2\sqrt{y}$. We obtain $${(1 + x)^2 \over y} + {(1 + y)^2 \over x} \geq 4 {x \over y} + 4{y \over x}$$ Use AM-GM one more time, getting ${x \over y} + {y \over x} \geq 2$. Substituting this in the above gives the result. (Note this is similar to what N.S. did.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.