Inequalities

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

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sqing

#1 h Apr 4, 2025, 3:52 AM

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Let $a,b,c> 0$ and $\frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$ . Prove that
$a+b+c\leq 3$

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#2 h Apr 5, 2025, 1:13 PM • 1 Y

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sqing wrote:

Let $a,b,c> 0$ and $\frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$ . Prove that
$a+b+c\leq 3$

Using the Cauchy-Schwarz inequality, we have:

$1 \leq \sum \frac{a}{a^2 + ab + c} = \sum \frac{a \left(1 + \frac{b}{a} + c\right)}{(a^2 + ab + c)\left(1 + \frac{b}{a} + c\right)}$
$\leq \sum \frac{a + b + ac}{(a + b + c)^2} = \frac{2(a + b + c) + (bc + ca + ab)}{(a + b + c)^2}$
$\Rightarrow (a + b + c)^2 \leq 2(a + b + c) + (bc + ca + ab) \leq 2(a + b + c) + \frac{1}{3}(a + b + c)^2$
$\Rightarrow \frac{2}{3}(a + b + c)^2 \leq 2(a + b + c) \Rightarrow a + b + c \leq 3$
Hence the inequality is proved. We have equality $\Leftrightarrow a = b = c = 1$ .

This post has been edited 1 time. Last edited by pooh123, Apr 5, 2025, 1:14 PM
Reason: typo

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sqing

#3 h Apr 5, 2025, 1:34 PM

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Very nice.Thanks.

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sqing

#4 h Apr 5, 2025, 2:00 PM

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Let $a,b,c> 0$ and $\frac{a+b+c}{3} \leq \frac{a}{2a +1} + \frac{b}{2b +1} + \frac{c}{2c +1} .$ Prove that
$\frac{a}{2a +1} + \frac{b}{2b +1} + \frac{c}{2c +1} \leq 1$ $ab+bc+ca\leq 3$

This post has been edited 1 time. Last edited by sqing, Apr 5, 2025, 2:13 PM

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#5 h Apr 16, 2025, 1:58 PM

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sqing wrote:

Let $a,b,c> 0$ and $\frac{a+b+c}{3} \leq \frac{a}{2a +1} + \frac{b}{2b +1} + \frac{c}{2c +1} .$ Prove that
$\frac{a}{2a +1} + \frac{b}{2b +1} + \frac{c}{2c +1} \leq 1$ $ab+bc+ca\leq 3$

Since
$\frac{a + b + c}{3} \leq \frac{a}{2a + 1} + \frac{b}{2b + 1} + \frac{c}{2c + 1},$ we have
$3 - \frac{2(a + b + c)}{3} \geq \frac{1}{2a + 1} + \frac{1}{2b + 1} + \frac{1}{2c + 1}.$
Let $t = a + b + c$ . Using the Cauchy-Schwarz inequality, we get:
$3 - \frac{2t}{3} \geq \frac{1}{2a + 1} + \frac{1}{2b + 1} + \frac{1}{2c + 1} \geq \frac{9}{2t + 3}.$
Multiplying both sides by $3(2t + 3)$ , we obtain:
$(9 - 2t)(2t + 3) \geq 9.$
Expanding the left-hand side:
$27 + 12t - 4t^2 \geq 27 \Rightarrow 4t(3 - t) \geq 0,$ which implies $t \leq 3$ .

Since $a + b + c \leq 3$ , we have:
$\frac{1}{2a + 1} + \frac{1}{2b + 1} + \frac{1}{2c + 1} \geq \frac{9}{2t + 3} \geq 1,$ so:
$\frac{a}{2a + 1} + \frac{b}{2b + 1} + \frac{c}{2c + 1} \leq 3 - 2 \cdot 1 = 1.$
Also, since
$(a + b + c)^2 \geq 3(ab + bc + ca),$ we get:
$9 \geq (a + b + c)^2 \geq 3(ab + bc + ca) \Rightarrow ab + bc + ca \leq 3.$

This post has been edited 2 times. Last edited by pooh123, Apr 16, 2025, 2:01 PM
Reason: typo

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sqing

#6 h Apr 16, 2025, 2:19 PM

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Let $a,b,c>0$ and $\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2} \geq 1$ .Prove that $ab+bc+ca\leq 3$ Let $a,b,c>0$ and $\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\geq 1$ .Prove that $a+b+c\leq 3$

This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 2:27 PM

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sqing

#7 h Apr 16, 2025, 2:20 PM

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Thanks.

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#8 h Apr 17, 2025, 3:01 PM

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sqing wrote:

Let $a,b,c>0$ and $\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\geq 1$ .Prove that $a+b+c\leq 3$

A good problem :thumbup:

:thumbup:

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sqing

#9 h Apr 17, 2025, 4:26 PM

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Good.Thanks.

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sqing

#10 h Apr 19, 2025, 1:05 PM

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Let $a,b\geq 0$ and $(a^3+1) (b^3+1)(a^3+b^3) \ge 2$ . Prove that
$a+b\geq 1$ Let $a,b\geq 0$ and $(a^3+1) (b^3+1)(a^3+b^3) \ge 72$ . Prove that
$a^2+a+2b \geq 4$

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DAVROS

#11 h Apr 20, 2025, 5:45 AM

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sqing wrote:

Let $a,b\geq 0$ and $(a^3+1) (b^3+1)(a^3+b^3) \ge 2$ . Prove that $a+b\geq 1$

solution

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sqing

#12 h Apr 20, 2025, 8:02 AM

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Very very nice.Thank DAVROS.

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