prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$

Question

Let $a,b,c$ be a sides of triangle

Such that : $a+b+c=1$ Then

prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$

My effort :

Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$

Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$

Also :

$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$

But I don't know how to complete it,any help is appreciated !

Answer 1

A standard trick to solve such problems is to use the substitution $a= y + z, b =x+z, c = x+y$, where $x, y, z>0$ (to see why this is so, draw the inscribed circle and pairs of equal tangents: $x, y, z$ are the lengths of the tangents)

So, we have $$x+y+z = \frac{1}{2}$$ and the inequality is $$ (x+y)^2 + (x+z)^2 + (y+z)^2 + 4 (x+y)(x+z)(y+z) < x+y+z$$

$$2x^2 + 2y^2 + 2z^2 + 2(xy+yz+xz) + 4 (x+y)(x+z)(y+z) < x+y+z $$

$$x^2 + y^2 +z^2 + (xy+yz+xz) + 2(x+y)(x+z)(y+z) < (x+y+z)^2 $$ $$ 2(x+y)(x+z)(y+z) < xy + xz +yz$$

$$(x+y)(x+z)(y+z) < (x+y+z) (xy+xz +yz)$$

$$ x^2y + xzy + x^2z + xz^2 + y^2x + y^2z + yxz + yz^2 < x^2y + x^2z + xyz + xy^2 +xyz +y^2z + xyz +xz^2 + yz^2 $$

$$2 xyz < 3xyz $$

Answer 2

Homogenizing the equation, we WTS

$$\frac{(a+b+c)^3}{2} -(a+b+c)(a^2+b^2+c^2)-4abc > 0$$

Expanding and factoring, we obtain

$$(a+b-c)(a-b+c)(-a+b+c) > 0$$

This is true via the triangle inequality.

Answer 3

we need to prove that $$2(a^2+b^2+c^2)+8abc<(a+b+c)^2$$ or $$8abc<\sum_{cyc}(2ab-a^2)$$ or $$8abc<\sum_{cyc}a(b+c-a)$$ or $$8abc<\sum_{cyc}a(b+c-a)(b+c+a)$$ or $$8abc<\sum_{cyc}a(b^2+c^2+2bc-a^2)$$ or $$2abc<\sum_{cyc}(a^2b+a^2c-a^3)$$ or $$2abc(a+b+c)<\sum_{cyc}a\sum_{cyc}(a^2b+a^2c-a^3)$$ or $$\sum_{cyc}2a^2bc<\sum_{cyc}(a^3b+a^3c+2a^2b^2+2a^2bc-a^4-a^3b-a^3c)$$ or $$\sum_{cyc}(2a^2b^2-a^4)>0,$$ which is $16S^2>0$ by the Heron's formula.

Answer 4

Using homogeneization and Ravi's substitution, the problem boils down to showing that

$$2(a+b+c)(a^2+b^2+c^2)+8abc \leq (a+b+c)^3 $$ holds for any triple $(a,b,c)$ of side lengths of a triangle, i.e. that

$$4(A+B+C)((B+C)^2+(A+C)^2+(A+B)^2)+8(A+B)(A+C)(B+C) \leq 8(A+B+C)^3 $$ holds for any triple $(A,B,C)$ of positive numbers. The previous inequality simplifies into $$ (A+B+C)(A^2+B^2+C^2+AB+AC+BC)+(A^2 B+B^2 A+A^2 C+C^2 A+B^2 C+C^2 B+2ABC) \leq (A+B+C)^3$$ which is equivalent to the trivial $-ABC\leq 0$.