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ALGEBRA INEQUALITY
Tony_stark0094   3
N Apr 23, 2025 by sqing
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
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Tony_stark0094
Apr 23, 2025
sqing
Apr 23, 2025
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ALGEBRA INEQUALITY
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Tony_stark0094
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#1 h  Apr 23, 2025, 12:17 AM • 2 Y
Y by PikaPika999, RainbowJessa
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
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Tony_stark0094
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#2 h  Apr 23, 2025, 12:35 AM • 2 Y
Y by PikaPika999, RainbowJessa
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Share some better sols.
This post has been edited 3 times. Last edited by Tony_stark0094, Apr 23, 2025, 12:53 AM
Reason: fjajja
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Sedro
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Sedro
#3 h  Apr 23, 2025, 12:54 AM • 1 Y
Y by kiyoras_2001
Add $\sum_{cyc}\tfrac{a(b+c)}{(b+c)} = \sum_{cyc} a$ to both sides to obtain \[\sum_{cyc}\frac{(a+b)(c+a)}{(b+c)} \ge 2(a+b+c).\]Substitute $x = a+b$, $y = b+c$, and $z=c+a$ to turn this into \[\sum_{cyc} \frac{xz}{y} \ge \sum_{cyc} x,\]which is clear by Muirhead.
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sqing
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sqing
#5 h  Apr 23, 2025, 5:26 AM
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Tony_stark0094 wrote:
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
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https://artofproblemsolving.com/community/c6h1555254p9481496
https://artofproblemsolving.com/community/c6h2486593p20910375
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