Question 1183930
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Let ax^2 + bx + c = 0 be a quadratic equation with no real roots and a,b > 0.
Prove that -b/(2a) > H, where H is the harmonic mean of the roots of this quadratic equation.
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<pre>
Let p and q be the roots of the given equation,


The harmonic mean  of the roots is


    H = {{{2/(1/p + 1/q)}}} = {{{(2pq)/(p+q)}}}.


According to Vieta's theorem,  pq = {{{c/a}}},  p+q = {{{-b/a}}},  so


    H = {{{(2pq)/(p+q)}}} = {{{(2*(c/a))/((-b/a))}}} = {{{-(2c)/b}}}.


Thus the inequality we need to prove takes the form


    {{{-b/(2a)}}} > {{{-(2c)/b}}}.


It is equivalent to


    {{{b/2a}}} < {{{(2c)/b}}}


which with positive "a" and "b" is equivalent to


    b^2 < 4ac,   or   b^2 - 4ac < 0.


The last inequality is equivalent to the condition that the given quadratic equation has no real roots.
</pre>

Proved and solved.