Inequalities

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High School Math Grades 9-12, Ages 13-18

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sqing

#1 h Mar 14, 2025, 8:09 AM

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Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3$ . Prove that
$a+b+a^2+b^2 \geq 28-6\sqrt{21}$ Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 4$ . Prove that
$a+b+a^2+b^2 \geq 10-4\sqrt 5$ Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5$ . Prove that
$a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9}$ Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 3+ \sqrt {5}$ . Prove that
$a+b+a^2+b^2 \geq 2$

This post has been edited 2 times. Last edited by sqing, Mar 14, 2025, 8:31 AM

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sqing

#2 h Mar 14, 2025, 8:48 AM

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Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq 4$ . Prove that
$a+b+a^2+ab+b^2 \geq \frac{27-11\sqrt {5}}{2}$ Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1}\geq \frac{2}{5}(7+ 2\sqrt {7})$ . Prove that
$a+b+a^2+ab+b^2 \geq 2$

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#3 h Mar 14, 2025, 11:02 AM

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sqing wrote:

Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 3$ . Prove that
$a+b+a^2+b^2 \geq 28-6\sqrt{21}$

WLOG $a \geq b$ .

If $a+b \geq 1$ , then
$a+b+a^2+b^2 \geq 1 \geq 28-6\sqrt{21}.$
If $a+b < 1$ , then
$\frac{1}{b^2-b+1} \leq \frac{1}{a^2-a+1}, a^2 + a + 1 \geq b^2 + b + 1$ which implies that
$\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1} \geq \frac{a^2+a+1}{b^2-b+1}+\frac{b^2+b+1}{a^2-a+1} \geq 3.$
Thus, by the Cauchy–Schwarz inequality,
$3 \geq 6-\left(\frac{a^2+a+1}{a^2-a+1}+\frac{b^2+b+1}{b^2-b+1}\right) =\frac{2(a-1)^2}{a^2-a+1}+\frac{2(b-1)^2}{b^2-b+1} \geq \frac{2(1-a+1-b)^2}{a^2-a+1+b^2-b+1}.$
Let $u = a+b$ and $v = ab$ . Then,
$3\left(u^2-u+1-2v\right) \geq 2\,(2-u)^2.$
Hence,
$u^2-4v \geq \frac{u^2-10u+4}{3}.$
Since
$a+b+a^2+b^2 = u^2-2v+u = \frac{u^2-4v}{2}+\frac{u^2}{2}+u,$ if $u \leq 5-\sqrt{21} < \frac{1}{2}$ , then
$\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2-10u+4}{6}+\frac{u^2}{2}+u =\frac{2}{3}\,(u^2-u+1).$
Because $u^2-u+1$ is monotonically decreasing on $u\in \left(0,5-\sqrt{21}\right]$ , it follows that
$\frac{2}{3}\,(u^2-u+1) \geq \frac{2}{3}\Bigl[(5-\sqrt{21})^2-(5-\sqrt{21})+1\Bigr] = 28-6\sqrt{21}.$
If $5-\sqrt{21} < u < 1$ , then since $u^2\geq 4v$ , we have
$\frac{u^2-4v}{2}+\frac{u^2}{2}+u \geq \frac{u^2}{2}+u \geq \frac{(5-\sqrt{21})^2}{2}+(5-\sqrt{21}) = 28-6\sqrt{21}.$
Therefore, in all cases,
$a+b+a^2+b^2 \geq 28-6\sqrt{21}.$

This post has been edited 5 times. Last edited by lbh_qys, Mar 14, 2025, 11:06 AM

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sqing

#4 h Mar 14, 2025, 1:43 PM

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Very very nice.Thank lbh_qys.

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DAVROS

#5 h Mar 16, 2025, 5:52 PM

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sqing wrote:

Let $a, b\geq 0$ and $\frac {a^2+a+1}{b^2-b+1}+\frac {b^2+b+1}{a^2-a+1} \geq 5$ . Prove that $a+b+a^2+b^2 \geq \frac {2(26-5\sqrt{13})}{9}$

solution

This post has been edited 2 times. Last edited by DAVROS, Mar 16, 2025, 9:24 PM

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sqing

#6 h Mar 17, 2025, 2:39 AM

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Very very nice.Thank DAVROS.

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sqing

#7 h Mar 17, 2025, 9:22 AM

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Let $a , b\geq 0$ and $\frac{1}{a^2+1}+\frac{1}{b^2+1} \le1-\frac{1 }{\sqrt2} .$ Show that $a+b+ab\geq5+4\sqrt 2$ Let $a , b\geq 0$ and $\frac{1}{a^2+1}+\frac{1}{b^2+1} \ge1+\frac{1 }{\sqrt2} .$ Show that $a+b+ab\leq1$ Let $a , b\geq 0$ and $\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{2(5+2\sqrt3) }{13} .$ Show that $a+b+ab\leq2$ Let $a , b\geq 0$ and $\frac{1}{a^2+1}+\frac{1}{b^2+1} \le \frac{15+\sqrt 5 }{10}.$ Show that $a+b+ab+a^2+b^2\geq 1$ Let $a , b$ be reals such that $\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{4(4+\sqrt3) }{13}.$ Show that $a+b+a^2+b^2\leq 1$ Let $a , b$ be reals such that $\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{9 }{5}.$ Show that $a+b+ab+a^2+b^2\leq 1$

This post has been edited 6 times. Last edited by sqing, Mar 17, 2025, 1:07 PM

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