Problem 305 Van Khea

September 1, 2011

Problem 305 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 2:33 am

If $a, b, c$ are positive real numbers such that $a+b+c=3$ . Prove that

$29(a^2+b^2+c^2)+30abc\geq 6(ab^2+bc^2+ca^2)+99$

Proof
Using the well-known identity $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca)$
Then the inequality above equivalent to
$29(ab+bc+ca)+3(ab^2+bc^2+ca^2)\leq 81+30abc$
$\Leftrightarrow 29.3(ab+bc+ca)+9(ab^2+bc^2+ca^2)\leq 243+90abc$
$\Leftrightarrow 29(a+b+c)(ab+bc+ca)+9(ab^2+bc^2+ca^2)\leq 243+90abc$
$\Leftrightarrow 29(a^2b+b^2c+c^2a)+38(ab^2+bc^2+ca^2)+42abc\leq 243$
Let $\displaystyle a=\frac{3x}{x+y+z}; b=\frac{3y}{x+y+z}; c=\frac{3z}{x+y+z}; \forall{x, y, z>0}$
The inequality transforms into
$29(x^2y+y^2z+z^2x)+38(xy^2+yz^2+zx^2)+42xyz\leq 9(x+y+z)^3$
$\Leftrightarrow 9(x^3+y^3+z^3)+12xyz\geq 3(x^2y+y^2z+z^2x)+11(xy^2+yz^2+zx^2)$
$\Leftrightarrow x(y+2z-3x)^2+y(z+2x-3y)^2+z(x+2y-3z)^2\geq 0$
Which is clear true. Equality occurs for $x=y=z\Leftrightarrow a=b=c=1$

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September 1, 2011