Let $a+b=2$, where $a\geq b>0$. Prove that: $$a^a(b+1)^2\leq b^b(a+1)^2$$
I tried to prove that $\frac{a^a}{(a+1)^2}\leq\frac{b^b}{(b+1)^2}$, but without success.
If $a+b=2$ and $a\geq b>0$ so $a^a(b+1)^2\leq b^b(a+1)^2$
4
$\begingroup$
calculus
real-analysis
inequality
contest-math
-
0If $a \in [1,2)$ and $b\in (0,1]$ then is $a/b\geq1$? – 2017-02-27
-
1Do you have the correct inequality symbol? I find that when $a=1.7$, the left side is $4.1653$ while the right side is $5.08$. – 2017-02-27
-
1This inequality seems to be incorrect - rather, it should be $a^a{(b+1)}^2 \le b^b{(a+1)}^2$ – 2017-02-27
-
1@ AnotherJohnDoe Let the OP edit his question when he come back. It will be in conflict with the author's intent if you do so. – 2017-02-27
1 Answers
4
With $a = 1+x, b = 1-x$ we have to show that $$ \frac{(1+x)^{1+x}}{(2+x)^2} \le \frac{(1-x)^{1-x}}{(2-x)^2} \quad \text{for } 0 \le x < 1 \, . $$ Taking logarithms this is equivalent to $$ \tag{*} (1+x)\log(1+x) - (1-x)\log(1-x) - 2 \log(2+x) +2 \log(2-x) \le 0\, . $$ Denoting the left-hand side with $h(x)$ we have $h(0) = 0$ and $$ h'(x) = \log(1+x) + \log(1-x) + 2 - \frac{2}{2+x} - \frac{2}{2-x} \\ = \log(1-x^2) - \frac{2x^2}{4 - x^2} < 0 $$ for $0 < x < 1$, which implies that $h$ is strictly decreasing. This completes the proof of $(*)$.
-
0Easy and convincing. Thank you! – 2017-02-27
-
0@MichaelRozenberg: You are welcome! – Your other question http://math.stackexchange.com/q/2161321/42969 looks much more difficult. – 2017-02-27
-
0This problem is driving me crazy – 2017-02-27
-
0Just by curiosity : considering $f=a^a(b+1)^2\- b^b(a+1)^2$ with, as you didv $a=1+x$ and $b=1-x$, the Taylor expansion (up to any order) built around $x=0$ only contains negative coefficients. – 2017-02-28