Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$

Question

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

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

Answer 1

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.

Answer 2

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 $x +\frac{1}{x} \geq 2$.

Answer 3

$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}=\frac{a^3-a^2+b^3-b^2}{(a-1)(b-1)}=\frac{(a+b)(a^2+b^2-ab)-(a^2+b^2)}{(a-1)(b-1)}\ge\frac{(a+b)ab-(a^2+b^2)}{(a-1)(b-1)}=\frac{a^2(b-1)+b^2(a-1)}{(a-1)(b-1)}=\frac{a^2}{a-1}+\frac{b^2}{b-1}=\frac{a^2-1+1}{a-1}+\frac{b^2-1+1}{b-1}=a+1+\frac{1}{a-1}+b+1+\frac{1}{b-1}=a-1+\frac{1}{a-1}+2+b-1+\frac{1}{b-1}+2\ge2+2+2+2=8$$

Answer 4

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

Answer 5

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 $f(2,2)=8$.

Answer 6

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, $$\frac{a^2}{a - 1} + \frac{b^2}{b - 1} = 4 + (a - 1) + \frac{1}{a - 1} + (b - 1) + \frac{1}{b - 1} \geq 4 + 2 + 2 = 8$$

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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.

Answer 7

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.)