shohvanilu wrote: Given a,b c are lenth of a triangle (If ABC is a triangle then AC=b, BC=a, AC=b) and $a+b+c=1$. Prove that $1+abc<ab+bc+ca\leq \frac{28}{27}+abc$ I have some grammar mistakes.Because my english is not well. But I think that you understand my problem. For the right inequality.. We know that $(a+b-c)(a-b+c)(b+c-a) \le abc$ for every $a, b, c$ are sides of a triangle. This gives $(1-2c)(1-2b)(1-2a) \le abc \Leftrightarrow 1-2(a+b+c)+4(ab+bc+ca) -9abc \le 0 \Leftrightarrow ab+bc+ca \le \frac{1}{4}+\frac{9}{4}abc \le abc + \frac{8}{27}<\frac{28}{27}+abc$. Since $abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27}$.

shohvanilu wrote: Given a,b c are lenth of a triangle (If ABC is a triangle then AC=b, BC=a, AC=b) and $a+b+c=1$. Prove that $1+abc<ab+bc+ca\leq \frac{28}{27}+abc$ I have some grammar mistakes.Because my english is not well. But I think that you understand my problem. For the left inequality.. It is not true because $ab+bc+ca \le \frac{1}{3}(a+b+c)^2 = \frac{1}{3} < 1 < 1+abc $.

shohvanilu wrote: Given a,b c are lenth of a triangle (If ABC is a triangle then AC=b, BC=a, AC=b) and $a+b+c=1$. Prove that $\frac{1}{4}+abc<ab+bc+ca\le \frac{8}{27}+abc$ I think this has to be the correct form.

The original problem with the corrected condition $a+b+c=2$ is quite nice. The right hand inequality, after homogenisation is Schur (and a,b,c don't need to be sides of a triangle) The left hand one is equivalent to $(a-1)(b-1)(c-1) < 0$ which is obvious since $a,b,c$ must all be $<1$ by the triangle inequality. Merlin

See here http://artofproblemsolving.com/community/q1h1156294p5487714

Let $x=\frac{-a+b+c}{2},y=\frac{a-b+c}{2},z=\frac{a+b-c}{2}$,then $x+y+z=1, x,y,z > 0$. So the inequality arrive at $(x+y+z)^3 < x^3+y^3+z^3+3(x^2y+ \dots z^2y)+7xyz \leq \frac{28}{27}(x+y+z)^3$ LHI $\Leftrightarrow xyz > 0$.This obviously holds. RHI $\Leftrightarrow 3S(3,0,0)+18S(2,1,0) \geq 21S(1,1,1)$.This holds by Muirhead inequality.

Prove that for all non-negative $a,b,c$ with $a+b+c=1$ the following inequality $$ab+bc+ca-2abc\le\frac{7}{27}$$holds. here