Let $a, b>1$ be reals such that $a+\frac{1}{a^2} \geq 5b-\frac{3}{b^2}$. Show that $a>5b-\frac{4}{b^2}$.
Problem
Source: St. Petersburg 2023 9.1
Tags: inequalities
sqing
12.08.2023 16:46
Let $a, b\geq1$ and $a+\frac{1}{a^2} \geq 5b-\frac{3}{b^2}$. Show that $$a\geq(k+1)b-\frac{k}{b^2}$$Where $0<k\leq 4$ here
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Tintarn
12.08.2023 17:03
a_507_bc wrote: Let $a, b>1$ be reals such that $a+\frac{1}{a^2} \geq 5b-\frac{3}{b^2}$. Show that $a>5b-\frac{4}{b^2}$.
Let $f(t)=t+\frac{1}{t^2}$. By assumption, $f(a) \ge 5b-\frac{3}{b^2}$. Moreover, we have $f\left(5b-\frac{4}{b^2}\right)<5b-\frac{3}{b^2}$. Indeed, this is equivalent to
\[\frac{1}{\left(5b-\frac{4}{b^2}\right)^2}<\frac{1}{b^2}\]and hence equivalent to $b<5b-\frac{4}{b^2}$ and hence to $b>1$ which is true by assumption.
Hence $f\left(5b-\frac{4}{b^2}\right)<f(a)$.
Finally, $f$ is increasing for $t \ge \sqrt[3]{2}$ and decreasing for $t<\sqrt[3]{2}$ so we are done unless $f(a) \le f(1)=2$.
But $f(a) \ge 5b-\frac{3}{b^2}>2+3b-\frac{3}{b^2}>2$ by assumption. Done.
grupyorum
12.08.2023 17:31
Suppose the contrary. Then, \[ 5b-\frac{4}{b^2}+\frac{1}{a^2}\ge a+\frac{1}{a^2}\ge 5b-\frac{3}{b^2}, \]yielding $b\ge a$. But then, \[ a+\frac{1}{a^2}\ge 5b-\frac{3}{b^2}\ge 5a-\frac{3}{a^2}\Rightarrow \frac{4}{a^2}\ge 4a, \]yielding $1\ge a$, a contradiction,
anhduy98
13.08.2023 02:31
For the collection. Given three real number $ k \ge \frac{1}{3} $ and $ a,b \ge 1 $ such that $ :a+\frac{1}{a^2} \ge (k+1)b-\frac{k-1}{b^2} $.Prove that :$$a \ge (k+1)b-\frac{k}{b^2}.$$ .