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.)

for this problem i have another solution a+b+c \leq 3 <=> \sum a ² + \sum 2ab \leq 9 <=> 4-abc+ \sum 2ab \leq 9 <=> 8 \sum ab - 9abc \leq 5 \sum a ² <=> (8 \sum ab - 5 \sum a ² )*(a+b+c) \leq 9abc*3 <=> \sum sym 5*a ^3 +abc - 6*a ² 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 ² +b ² +c ² +d ² +abcd=5. prove that a+b+c+d \leq 4. what do you think?

For positive $a,b,c$, and $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 For positive $a,b,c$, and $ a^2+b^2+c^2+abc=4$, prove that $$a+b+c\leq 3$$$$abc\leq 1$$$$a+b+c\geq 2+abc$$$$a+b+c=2+\sqrt{(2-a)(2-b)(2-c)}$$$$ab+bc+ca\leq 2+abc\iff cosAcosB+cosBcosC+cosCcosA\leq \frac{1}{2}+2cosAcosBcosC$$$$abc\leq(2-a)(2-b)(2-c)\iff cosAcosBcosC\leq(1-cosA)(1-cosB)(1-cosC)$$$$a+b+c\geq 2+\sqrt{abc}$$$$a+b+c\geq abc+2\sqrt{abc}$$$$a(b+3)+b(c+3)+c(a+3)\geq 6+abc+5\sqrt{abc}$$$$2(a+b+c)\leq 4+ab+bc+ca-abc\leq 6$$$$4(a+b+c)\leq (a+b)^2+(b+c)^2+(c+a)^2\leq 12$$

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define $f=3-a-b-c$ ,$g=a^2+b^2+c^2-abc-4=0$ $L=f-\lambda g$ . $\frac{\partial L}{\partial a}=-1-\lambda(2a-bc)=0$ 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 , $f_{min}=0$. hence done

2nd one from here

sqing wrote: For positive $a,b,c$, and $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 Set $a=\dfrac{2x}{z+y},b=\dfrac{2y}{x+z},c=\dfrac{2z}{x+y}4$ => Nessbit inequality

Omid Hatami wrote: For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\] Prove $a+b+c \leq3$ For $a+b+c=3$. Prove that: $a^2+b^2+c^2+abc \le 4$

Omid Hatami wrote: For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\] Prove $a+b+c \leq3$ $a ^2+abc+b^2+c^2-4=0 $ Taking this as a quadratic in $a$ we have $a=\frac{-bc+\sqrt{(4-b^2)(4-c^2)}}{2} $ Applying Am-Gm we get: $a \le \frac{-2bc+(4-b^2)+(4-c^2)}{4} $ $a+b+c\le \frac{-2bc+(b^2+c^2-8+4b+4c)}{4}=\frac{12-(b+c-2)^2}{4}\le 3$ Sayan wrote: 2nd one from here I'd like to see the solution for the 6th problem.

sqing wrote: For positive $a,b,c$, and $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251 For a stronger inequality, see Vietnam MO in 1996

USA MO $ab+bc+ac \leq abc+2$ if $a,b,c\leq 0,a^2+b^2+c^2+abc=4$ Set a=2cosA,b=2cosB,c=2cosC=>A,B,C are 3 angles of an acute triangle $<=> 4\sum cosA.cosB\leq 8cosA.cosB.cosC+2$ $<=>4\sum cosA.cosB\leq 4(1-\sum cos^{2}A)+2$ $<=>(\sum cosA)^{2}\leq \sum sin^{2}A$ +> I think you can prove this inequality

We denote by : $ a=\frac{2x}{y+z},b=\frac{2y}{z+x},c=\frac{2z}{x+y} $ , then we have : $ ab+bc+ca+abc=4\Leftrightarrow \sum \frac{4xy}{(y+z)(z+x)}+\frac{8xyz}{\prod (x+y)}=4\Leftrightarrow \sum xy(x+y)+2xyz=\prod (x+y) $ ,it's true , we : $ a+b+c=2(\sum \frac{x}{y+z})\geq(Nesbitt's) 3 $

$ we~know~that~a+(\frac{b+c}{2})^2\le 2~thus~b+c \le 2\sqrt{2-a}~\\ thus~a+b+c \le a+2\sqrt{2-a}\le a+(2-a)+1=3 $ http://www.artofproblemsolving.com/blog/78923

We have: $ a,b,c >0,a^2+b^2+c^2+abc=4\Leftrightarrow \sum\frac{bc}{2a+bc}= 1 \Leftrightarrow \sum\frac{b^2c^2}{2abc+b^2c^2}= 1$ Applying Cachy-Scwarz get: $ 1=\sum\frac{b^2c^2}{2abc+b^2c^2}\geq\frac{(bc+ca+ab)^2}{6abc+b^2c^2+c^2a^2+a^2b^2}\Rightarrow 6abc+b^2c^2+c^2a^2+a^2b^2\geq (bc+ca+ab)^2\Leftrightarrow 6abc\geq 2abc(a+b+c)\Leftrightarrow 3\geq a+b+c $ Q.E.D ________________ Sandu Marin

Let $a,b,c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}+abc\le 4$ .Prove that\[a+b+c \leq3.\] Let $a,b,c$ be positive real numbers satisfying $a^{3}+b^{3}+c^{3}+abc\le 4$ .Prove that\[a+b+c \leq3.\] Let $a,b,c$ be positive real numbers satisfying $a^{5}+b^{5}+c^{5}+abc\le 4$ .Prove that\[a+b+c \leq3.\] Let $a,b,c$ be positive real numbers satisfying $ a+b+c=3 $. Prove that\[ a^3+b^3+c^3+abc\ge 4.\] 《Mathematics teaching》(China Shanghai)No.10(2013)problem 897: Let $a,b,c$ be nonnegative real numbers satisfying $ a+b+c=3 $. Prove that\[9\ge a^2+b^2+c^2+abc\ge 4.\]

For positive $a,b$ such that $a^{2}+b^{2}+ab=3 $ , Prove that \[a+b \leq2.\]proof

For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\]

sqing wrote: Let $a,b,c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}+abc\le 4$ .Prove that\[a+b+c \leq3.\] We have: $9=(a^2+b^2+c^2+2abc+1)+a^2+b^2+c^2 \ge 2(ab+bc+ac)+a^2+b^2+c^2=(a+b+c)^2$ Then $\Rightarrow 3\ge a+b+c$.

sqing wrote: For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\] Contradiction method kills it.

sqing wrote: For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\] See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=123037

Strengthening For nonnegative real numbers $a,b,c$ satisfying $a^{2}+b^{2}+c^{2}+\frac{3}{2}abc = \frac{9}{2}$. Prove that \[a+b+c \leq3.\]

sqing wrote:

For positive $a,b$ such that $a^{2}+b^{2}+ab=3 $ , Prove that \[a+b \leq2.\]proof

$a+b=\sqrt{\frac{4(a+b)^2}{3}-\frac{(a+b)^2}{3}}\le\sqrt{\frac{4(a+b)^2}{3}-\frac{4ab}{3}}=2.$

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 $a=b=c=...$

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.

sqing wrote: sqing wrote:

For positive $a,b$ such that $a^{2}+b^{2}+ab=3 $ , Prove that \[a+b \leq2.\]proof

$a+b=\sqrt{\frac{4(a+b)^2}{3}-\frac{(a+b)^2}{3}}\le\sqrt{\frac{4(a+b)^2}{3}-\frac{4ab}{3}}=2.$ 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. $\Rightarrow a+b\leqslant 2$

Omid Hatami wrote: For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\]Prove $$a+b+c \leq3$$ Generalization:

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive real numbers . Prove that $$a_1^2+a_2^2+\cdots+a_n^2+a_1a_2\cdots a_n+2n-1\geq 3(a_1+a_2+\cdots+a_n).$$n=3

Let $a,b,c$ be reals such that $ a^3+b^3+c^3+2abc=5.$ Prove that $$ a+b+c\leq 3$$Let $a,b,c$ be reals such that $ a^4+b^4+c^4+2abc=5.$ Prove that $$ a+b+c\leq 3$$Let $a,b,c,d$ be reals such that $a^4+b^4+c^4+d^4+2abcd=6.$ Prove that $$ a+b+c+d \leq 4$$Let $a,b,c$ 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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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 \[a^{3}+b^{3}+c^{3}+abc+8\geq 4(a+b+c)\]

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 that$$a^3+b^3+c^3+5a^2b^2c^2\le 8$$For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ . Prove that$$a^3+b^3+c^3+d^3+a+b+c+d\geq 8$$For positive $a,b,c$ such that $a^{2}+b^{2}+c^{2}+abc=4 .$ Prove that$$a^3+b^3+c^3+ab+bc+ca\geq 6$$

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