An inequality

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0 replies

jlacosta

Apr 2, 2025

0 replies

inequalities 070425

pennypc123456789 5

N 3 hours ago by Sadigly

Let $a,b,c$ be positive real numbers . Prove that :
$\dfrac{2ab}{a^2+b^2} + \dfrac{2bc}{b^2+c^2} + \dfrac{2ac}{a^2+c^2} \ge \dfrac{24abc}{(a+b)(b+c)(a+c)}$

5 replies

pennypc123456789

5 hours ago

Sadigly

3 hours ago

inequalities such a hard problem

mannothot11 10

N 5 hours ago by sqing

for all real x y z and x^2 + y^2 +z^2 = 1
Prove that xy + y^2 +2xz≤ (√3 +1)/2

ps: forgive me because i don know how to edit this

10 replies

mannothot11

Jan 16, 2018

sqing

5 hours ago

Inequalities

sqing 6

N 5 hours ago by sqing

Let $a,b$ be real numbers such that $a^2+b^2+a^3 +b^3=4 .$ Prove that
$a+b \leq 2$ Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 .$ Prove that
$a+b \leq 2$

6 replies

sqing

Saturday at 1:10 PM

sqing

5 hours ago

Congruence

Ecrin_eren 3

N Today at 3:39 AM by lbh_qys

Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.

3 replies

Ecrin_eren

Apr 3, 2025

lbh_qys

Today at 3:39 AM

No more topics!

An inequality

JK1603JK 2

N Apr 2, 2025 by lbh_qys

Let a,b,c\ge 0: a+b+c=3 then prove \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{2}\cdot\frac{1}{2ab+2bc+2ca+3}.

2 replies

JK1603JK

Apr 2, 2025

lbh_qys

Apr 2, 2025

An inequality

G H J

Reveal topic tags

inequalities

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JK1603JK

44 posts

JK1603JK

#1 h Apr 2, 2025, 3:11 AM

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Let a,b,c\ge 0: a+b+c=3 then prove \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{2}\cdot\frac{1}{2ab+2bc+2ca+3}.

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aidan0626

1819 posts

aidan0626

#2 h Apr 2, 2025, 3:21 AM

Y by

JK1603JK wrote:

Let $a,b,c\ge 0$ : $a+b+c=3$ then prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le \frac{27}{2}\cdot\frac{1}{2ab+2bc+2ca+3}$ .

Z K Y

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lbh_qys

482 posts

lbh_qys

#3 h Apr 2, 2025, 3:24 AM

Y by

Let $q = ab+bc+ca$ and $r = abc$ ; then, it is equivalent to proving that

$\frac{q+9}{r+q+4} \leq \frac{27}{4q+6}$
i.e.,

$27r \geq (q+9)(4q+6) - 27(q+4) = (4q-9)(q+6).$
According to Schur’s inequality, $4q-9 \leq 3r$ , and furthermore, $q+6 \leq 9$ ; hence, the above inequality obviously holds, thereby establishing the original inequality.

Z K Y

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