Inequality 4

[MM]

For positive numbers a, b and c prove that:
[tex]\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc}\le \frac{1}{abc}[/tex]

I have found nice solution.

[Rock'n'roller]

Well mine isn't great but I think it's true

The inequality is [tex]\Leftrightarrow[/tex] with

[tex]a^3b^6 + b^3c^6 + c^3a^6 + a^6b^3 + b^6c^3 + c^6a^3 \ge 2a^5b^2c^2 + 2a^2b^5c^2 + 2a^2b^2c^5[/tex]

But (1) [tex]x^3 + y^3 + z^3 \ge x^2y + y^2z + z^2x \Leftrightarrow (\frac{x^3+x^3+y^3-3x^2y}{3}) + (\frac{y^3+y^3+z^3-3y^2z}{3}) + (\frac{z^3+z^3+x^3-3z^2x}{3}) \ge 0[/tex] so (1) is true. If we change x with y, y with z, z with x, we will find that [tex]x^3 + y^3 + z^3 \ge xy^2 + yz^2 + zx^2[/tex] (2). Now, using (1) we obtain

[tex](ab^2)^3 + (bc^2)^3 + (ca^2)^3 \ge (ab^2)^2.bc^2 + (bc^2)^2.ca^2 + (ca^2)^2.ab^2[/tex] or [tex]a^3b^6 + b^3c^6 + c^3a^6 \ge a^5b^2c^2 + a^2b^5c^2 + a^2b^2c^5[/tex] (1*) and using (2) - [tex]a^6b^3 + b^6c^3 + c^6a^3 \ge a^5b^2c^2 + a^2b^5c^2 + a^2b^2c^5[/tex] (2*). Then we just add (1*) to (2*) and all is OK(maybe?).

[MM]

Great solution Rock'n'roller!
Mine is the following:
Assume [tex]abc=1[/tex] . Let [tex]a=\frac{1}{\sqrt[3]{x}}[/tex], [tex]b=\frac{1}{\sqrt[3]{b}}[/tex], [tex]c=\frac{1}{\sqrt[3]{z}}[/tex]. Then the inequality is equivalent to
[tex]\frac{xy}{x+y+xy}+\frac{zx}{z+x+zx}+\frac{yz}{y+z+yz}\le 1[/tex]. After expanding and using the fact [tex]xyz=1[/tex] we obtain [tex]x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y\ge 2\left(xy+zx+yz\right)[/tex]. Assume [tex]x+y+z=p[/tex] and [tex]xy+xz+yz=q[/tex]. Then using the fact [tex]xyz=1[/tex] we obtain [tex]pq-3\ge 2q \Leftrightarrow q\left(r-2\right)\ge 3[/tex]. Using AM-GM we obtain [tex]p\ge 3[/tex] and [tex]q\ge 3[/tex]. So the last we had to prove is done.

[Rock'n'roller]

Thanks but please explain this

Assume [tex]abc=1[/tex].

[MM]

Since the inequality is in tree variables and homogeneous we may assume [tex]a+b+c=k[/tex], [tex]ab+ca+bc=k[/tex] or [tex]abc=k[/tex] where "k" is a number which doesn't contradicts with any conditions for a,b and c (for example if it is given [tex]a,b,c>0[/tex] we can't assume [tex]ab+ca+bc=-1[/tex]).

[MM]

The inequality is [tex]\Leftrightarrow[/tex] with

[tex]a^3b^6 + b^3c^6 + c^3a^6 + a^6b^3 + b^6c^3 + c^6a^3 \ge 2a^5b^2c^2 + 2a^2b^5c^2 + 2a^2b^2c^5[/tex]

I think this follows directly from the Muirhead's inequality.

[Rock'n'roller]

Which is...?

[poster123]

Thanks but please explain this

[Guest]

Since the inequality is in tree variables and homogeneous we may assume a+b+c=k, ab+ca+bc=k or abc=k where "k" is a number which doesn't contradicts with any conditions for a,b and c (for example if it is given a,b,c>0 we can't assume ab+ca+bc=-1

[sajid121]

Since the inequality is in tree variables and homogeneous we may assume a+b+c=k, ab+ca+bc=k or abc=k where "k" is a number which doesn't contradicts with any conditions for a,b and c (for example if it is given a,b,c>0 we can't assume ab+ca+bc=-1).