Let $a$ be a real number and $b$ a real number with $b\neq-1$ and $b\neq0. $ Find all pairs $ (a, b)$ such that $$\frac{(1 + a)^2 }{1 + b}\leq 1 + \frac{a^2}{b}.$$For which pairs (a, b) does equality apply? (Walther Janous)
Problem
Source: 2020 Austrian Mathematical Olympiad Junior Regional Competition , Problem 1
Tags: algebra, inequalities, High school olympiad, Austria
SomeUser221104
07.06.2020 12:19
This a result of Titu's Lemma, thus equality happens iff $a=b$. Thus the pairs are,
$$ \{(k,k) | k \neq 0,-1 , k \in \mathbb{R} \} $$
Equality happens when $\frac{a^2+b}{b} = \frac{a^2+2a+1}{b+1} $, using the identity $\frac{x}{y} = \frac{a}{b}=\frac{a-x}{b-y} $, we have
$$ \frac{a^2+b}{b} =2a+1-b \Leftrightarrow (a-b)^2= 0$$Thus equality happens iff. $a=b$.
sqing
07.06.2020 12:23
SomeUser221104 wrote:
The Straightforward SolutionThis a result of Titu's Lemma, thus equality happens iff $a=b$. Thus the pairs are,
$$ \{(k,k) | k \neq 0,-1 , k \in \mathbb{R} \} $$
Another SolutionEquality happens when $\frac{a^2+b}{b} = \frac{a^2+2a+1}{b+1} $, using the identity $\frac{x}{y} = \frac{a}{b}=\frac{a-x}{b-y} $, we have
$$ \frac{a^2+b}{b} =2a+1-b \Leftrightarrow (a-b)^2= 0$$Thus equality happens iff. $a=b$.
arzhang2001
07.06.2020 12:38
it just C-S: $(1 + a)^2 =({1 + b})(1 + \frac{a^2}{b}).$
test20
07.06.2020 13:08
By using common denominators and by bringing everything to the same side of the inequality sign, we get the equivalent inequality $$ \frac{(a-b)^2}{b(b-1)} ~\ge~ 0. $$ Since the numerator is non-negative, the inequality holds if and only if the numerator equals $0$ (which means $a=b$) or if the denominator is positive (which means $b<0$ or $b>1$). Hence the answer to the problem is: All pairs $(a, b)$ with $a=b$ or $b<0$ or $b>1$. And equality holds if and only if $a=b$.
Ln142
07.06.2020 17:57
Nice inequality, I solved it yesterday when the problems got posted on the online page. Do you speak German? My solution:
We can look at two separate cases:
Case 1: $b<-1$ or $b>0$
Now, we can multiply with $b$ and $b+1$ without having to change the direction of the inequality
We get:
$\left({1+a}\right)^2b \leq b(b+1) +a^2(b+1)$
$2ab \leq a^2+b^2 \iff$
$\left({a-b}\right)^2 \geq 0 \iff$
Equality applies if $a=b$, for all Pairs $(a,a)$
Case 2: $ -1<b<0$
We get:
$\left({a-b}\right)^2 \leq 0$
This inequality is obviously not true. (except for $a=b$)
So the answer is all Pairs $(a,b)$ with $b<-1$ or $ b>0$ ; $b\neq0,-1$ and equality for all pairs $(a, a)$ ; $a\neq0,-1$
sqing
08.06.2020 05:01
Ln142 wrote:
Nice inequality, I solved it yesterday when the problems got posted on the online page. Do you speak German?
My solution:
Click to reveal hidden textWe can look at two separate cases:
Case 1: $b<-1$ or $b>0$
Now, we can multiply with $b$ and $b+1$ without having to change the direction of the inequality
We get:
$\left({1+a}\right)^2b \leq b(b+1) +a^2(b+1)$
$2ab \leq a^2+b^2 \iff$
$\left({a-b}\right)^2 \geq 0 \iff$
Equality applies if $a=b$, for all Pairs $(a,a)$
Case 2: $ -1<b<0$
We get:
$\left({a-b}\right)^2 \leq 0$
This inequality is obviously not true. (except for $a=b$)
So the answer is all Pairs $(a,b)$ with $b<-1$ or $ b>0$ ; $b\neq0,-1$ and equality for all pairs $(a, a)$ ; $a\neq0,-1$
Sadigly
11.11.2024 00:14
1)$b(b+1)>0$ $b(1+a)^2\leq b(b+1)+a^2(b+1)$ $a^2b+2ab+b\leq b^2+b+a^2b+a^2$ $2ab\leq b^2+a^2$ which is true 2)$b(b+1)<0$ $2ab\geq b^2+a^2$ which is true if and only if $a=b$ So, this inequality is always true for $b\in(-\infty;-1)\cup(0;\infty)$, and this equality holds if $a=b$ for $b\in(-1;0)$
AshAuktober
19.11.2024 05:38
This is simply Titu's lemma.