For positive a,b,c, a2+b2+c2+abc=4 Prove a+b+c≤3
Problem
Source: Iran selection test 2002
Tags: quadratics, Iran, Inequality, algebra
03.04.2004 21:04
Put a = 2 cos A, b = 2 cos B and c = 2 cos C where a + b + c = \pi. Then use the fact that R \geq 2r and cos A + cos B + cos C = 1 + r/R in any triangle. (You can use Jensen too, if you want, or if you don't know the above identity.)
04.04.2004 08:03
for this problem i have another solution a+b+c \leq 3 <=> \sum a <sup>2</sup> + \sum 2ab \leq 9 <=> 4-abc+ \sum 2ab \leq 9 <=> 8 \sum ab - 9abc \leq 5 \sum a <sup>2</sup> <=> (8 \sum ab - 5 \sum a <sup>2</sup> )*(a+b+c) \leq 9abc*3 <=> \sum sym 5*a ^3 +abc - 6*a <sup>2</sup> b \geq 0 by using schur we have a+b+c \leq 3 by using this solution i think we can solve the problem let a <sup>2</sup> +b <sup>2</sup> +c <sup>2</sup> +d <sup>2</sup> +abcd=5. prove that a+b+c+d \leq 4. what do you think?
31.01.2013 04:52
For positive a,b,c, and ab+bc+ca+abc=4, prove that a+b+c≥3.http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 For positive a,b,c, and a2+b2+c2+abc=4, prove that a+b+c≤3abc≤1a+b+c≥2+abca+b+c=2+√(2−a)(2−b)(2−c)ab+bc+ca≤2+abc⟺cosAcosB+cosBcosC+cosCcosA≤12+2cosAcosBcosCabc≤(2−a)(2−b)(2−c)⟺cosAcosBcosC≤(1−cosA)(1−cosB)(1−cosC)a+b+c≥2+√abca+b+c≥abc+2√abca(b+3)+b(c+3)+c(a+3)≥6+abc+5√abc2(a+b+c)≤4+ab+bc+ca−abc≤64(a+b+c)≤(a+b)2+(b+c)2+(c+a)2≤12
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31.01.2013 13:28
define f=3−a−b−c ,g=a2+b2+c2−abc−4=0 L=f−λg . ∂L∂a=−1−λ(2a−bc)=0 etc. so , we see that 2a−bc=2b−ca=2c−ab so ,now , we see that a=b=c . so a=b=c=1 . so , fmin=0. hence done
31.01.2013 15:08
2nd one from here
31.01.2013 16:25
sqing wrote: For positive a,b,c, and ab+bc+ca+abc=4, prove that a+b+c≥3. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 Set a=2xz+y,b=2yx+z,c=2zx+y4 => Nessbit inequality
31.01.2013 16:56
Omid Hatami wrote: For positive a,b,c, a2+b2+c2+abc=4 Prove a+b+c≤3 For a+b+c=3. Prove that: a2+b2+c2+abc≤4
31.01.2013 17:21
Omid Hatami wrote: For positive a,b,c, a2+b2+c2+abc=4 Prove a+b+c≤3 a2+abc+b2+c2−4=0 Taking this as a quadratic in a we have a=−bc+√(4−b2)(4−c2)2 Applying Am-Gm we get: a≤−2bc+(4−b2)+(4−c2)4 a+b+c≤−2bc+(b2+c2−8+4b+4c)4=12−(b+c−2)24≤3 Sayan wrote: 2nd one from here I'd like to see the solution for the 6th problem.
31.01.2013 17:53
sqing wrote: For positive a,b,c, and ab+bc+ca+abc=4, prove that a+b+c≥3. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 For a stronger inequality, see Vietnam MO in 1996
31.01.2013 17:56
USA MO ab+bc+ac≤abc+2 if a,b,c≤0,a2+b2+c2+abc=4 Set a=2cosA,b=2cosB,c=2cosC=>A,B,C are 3 angles of an acute triangle <=>4∑cosA.cosB≤8cosA.cosB.cosC+2 <=>4∑cosA.cosB≤4(1−∑cos2A)+2 <=>(∑cosA)2≤∑sin2A +> I think you can prove this inequality
31.01.2013 19:51
We denote by : a=2xy+z,b=2yz+x,c=2zx+y , then we have : ab+bc+ca+abc=4⇔∑4xy(y+z)(z+x)+8xyz∏(x+y)=4⇔∑xy(x+y)+2xyz=∏(x+y) ,it's true , we : a+b+c=2(∑xy+z)≥(Nesbitt′s)3
01.02.2013 07:11
we know that a+(b+c2)2≤2 thus b+c≤2√2−a thus a+b+c≤a+2√2−a≤a+(2−a)+1=3 http://www.artofproblemsolving.com/blog/78923
05.02.2013 14:54
We have: a,b,c>0,a2+b2+c2+abc=4⇔∑bc2a+bc=1⇔∑b2c22abc+b2c2=1 Applying Cachy-Scwarz get: 1=∑b2c22abc+b2c2≥(bc+ca+ab)26abc+b2c2+c2a2+a2b2⇒6abc+b2c2+c2a2+a2b2≥(bc+ca+ab)2⇔6abc≥2abc(a+b+c)⇔3≥a+b+c Q.E.D ________________ Sandu Marin
24.10.2013 09:21
Let a,b,c be positive real numbers satisfying a2+b2+c2+abc≤4 .Prove thata+b+c≤3. Let a,b,c be positive real numbers satisfying a3+b3+c3+abc≤4 .Prove thata+b+c≤3. Let a,b,c be positive real numbers satisfying a5+b5+c5+abc≤4 .Prove thata+b+c≤3. Let a,b,c be positive real numbers satisfying a+b+c=3. Prove thata3+b3+c3+abc≥4. 《Mathematics teaching》(China Shanghai)No.10(2013)problem 897: Let a,b,c be nonnegative real numbers satisfying a+b+c=3. Prove that9≥a2+b2+c2+abc≥4.
18.02.2014 02:59
For positive a,b,c,d such that a2+b2+c2+d2+abcd=5 , Prove that a+b+c+d≤4.
20.02.2014 08:42
sqing wrote: Let a,b,c be positive real numbers satisfying a2+b2+c2+abc≤4 .Prove thata+b+c≤3. We have: 9=(a2+b2+c2+2abc+1)+a2+b2+c2≥2(ab+bc+ac)+a2+b2+c2=(a+b+c)2 Then ⇒3≥a+b+c.
21.02.2014 19:04
sqing wrote: For positive a,b,c,d such that a2+b2+c2+d2+abcd=5 , Prove that a+b+c+d≤4. Contradiction method kills it.
03.06.2014 20:39
sqing wrote: For positive a,b,c,d such that a2+b2+c2+d2+abcd=5 , Prove that a+b+c+d≤4. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=123037
05.01.2015 09:37
Strengthening For nonnegative real numbers a,b,c satisfying a2+b2+c2+32abc=92. Prove that a+b+c≤3.
06.01.2015 06:04
sqing wrote:
a+b=√4(a+b)23−(a+b)23≤√4(a+b)23−4ab3=2.
06.01.2015 06:12
sqing wrote: Strengthening For nonnegative real numbers a,b,c satisfying a2+b2+c2+32abc=92. Prove that a+b+c≤3. Assume a+b+c>3. We have: 9=18(a2+b2+c2)9+81abc27>18(a2+b2+c2)(a+b+c)2+81abc(a+b+c)3 ⇔2(ab+bc+ca)>a2+b2+c2+9abca+b+c is false. So a+b+c⩽ and equality when a=b=c=...
06.01.2015 06:37
dogsteven wrote: sqing wrote: Strengthening For nonnegative real numbers a,b,c satisfying a^{2}+b^{2}+c^{2}+\frac{3}{2}abc = \frac{9}{2}. Prove that a+b+c \leq3. Assume a+b+c>3. We have: 9=\dfrac{18(a^2+b^2+c^2)}{9}+\dfrac{81abc}{27}> \dfrac{18(a^2+b^2+c^2)}{(a+b+c)^2}+\dfrac{81abc}{(a+b+c)^3} \Leftrightarrow 2(ab+bc+ca)>a^2+b^2+c^2+\dfrac{9abc}{a+b+c} is false. So a+b+c\leqslant 3 and equality when a=b=c=... Nice. Thanks.
06.01.2015 08:22
sqing wrote: sqing wrote:
a+b=\sqrt{\frac{4(a+b)^2}{3}-\frac{(a+b)^2}{3}}\le\sqrt{\frac{4(a+b)^2}{3}-\frac{4ab}{3}}=2. If a+b>2 then 3=\dfrac{4(a^2+b^2)}{4}+\dfrac{4ab}{4}> \dfrac{4(a^2+b^2)}{(a+b)^2}+\dfrac{4ab}{(a+b)^2} \Leftrightarrow (a+b)^2>2(a^2+b^2) is false. \Rightarrow a+b\leqslant 2
09.05.2019 16:20
Omid Hatami wrote: For positive a,b,c, a^{2}+b^{2}+c^{2}+abc=4Prove a+b+c \leq3 Generalization:
01.09.2021 02:53
Let a,b,c be reals such that a^3+b^3+c^3+2abc=5. Prove that a+b+c\leq 3Let a,b,c be reals such that a^4+b^4+c^4+2abc=5. Prove that a+b+c\leq 3Let a,b,c,d be reals such that a^4+b^4+c^4+d^4+2abcd=6. Prove that a+b+c+d \leq 4Let a,b,c be nonnegative real numbers.Prove that a^3+b^3+c^3+4abc+4\geq 3(a+b+c)Let a,b,c,d be nonnegative real numbers.Prove that a^3+b^3+c^3+d^3+5abc+6\geq 3(a+b+c+d)Let a,b,c\in[1.2]. Prove that 1\leq a^4+b^4+c^4-2abc\leq 32 sqing wrote: Let a,b,c be positive real numbers .Prove a^2+b^2+c^2+abc+5\geq 3(a+b+c).
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01.12.2022 04:37
Let a,b,c be positive real numbers .Prove a^2+b^2+c^2+kabc+2k+3\geq (k+2)(a+b+c)Where k\in [0,\sqrt 2]. sqing wrote: Let a,b,c be positive real numbers .Prove a^2+b^2+c^2+abc+5\geq 3(a+b+c). https://artofproblemsolving.com/community/c4h1999552p13979821 https://artofproblemsolving.com/community/c6h2045260p14490458 Let a,b,c be positive real numbers .Prove that a^{3}+b^{3}+c^{3}+abc+8\geq 4(a+b+c)
11.02.2023 12:12
sqing wrote: For positive a,b,c,d such that a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 , Prove that a+b+c+d \leq4. For a,b,c>0 and a^2+b^2+c^2+abc=4. Prove thata^3+b^3+c^3+5a^2b^2c^2\le 8For positive a,b,c,d such that a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 . Prove thata^3+b^3+c^3+d^3+a+b+c+d\geq 8For positive a,b,c such that a^{2}+b^{2}+c^{2}+abc=4 . Prove thata^3+b^3+c^3+ab+bc+ca\geq 6
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