\hspace{-16}$Find Max. of Constant $k$ such that for any positive real no. \\\\ $a\;b\;,c$ with $abc=1$ satisfy the Inequality\\\\ $a^2+b^2+c^2+3k\geq (k+1)(ab+bc+ca)$
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1 Answers
62x^3-px^2+qx-1=0 \\p^2-2q+3k\ge (k+1)q \\p^2-3q\ge k(q-3)
q>3 bcos of am gm inequality (except when a=b=c.. then q=0 and then there is nothing to prove in this question!)
so we can divide by q-1
\\\frac{p^2-3q}{q-3}\ge k \\\frac{p^2-3q+9-9}{q-3}\ge k \\\frac{p^2-9}{q-3}-3\ge k
3x^2-2px+q has real roots
so 4p^2-12q>0 so p^2-3q>0
Thsi din help :(
But cos i typed so much (so i better posted this! :/)
There is one thing possible but that seems so cumbersome to try f(t)f(t') <0 where t, t' are the roots of the equation 3x^2-2px+q
