Solving an equality in 2 variables

Question

I need to prove that

$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$

if $a+b = 1$ and $a b \le 1/4$

I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get $a b \le 4$ in the inequality ?

Wouldn't it be a pro to have a preview of your answer you can check your Latex formulas ? — Ignace, Mar 17 '14 at 12:31
@Ignace the preview of the answer does show latex-versions of the text, you just have to use it (dollar signs and all) — 5xum, Mar 17 '14 at 12:39
A preview would prevent you from going in and out with the edit function to correct errors, so that If it looks pretty good you submit. Sure The answer gives the latex if it's there. — Ignace, Mar 17 '14 at 12:44
$;a=3,,,b=-2;$ is another counterexample...out of many more. — DonAntonio, Mar 17 '14 at 12:45
@Ignace for feature proposals, try meta.math.stackexchange.com. — 5xum, Mar 17 '14 at 12:49
http://math.stackexchange.com/questions/487486/proving-inequality-a-frac1a2-b-frac1b2-geq-frac252-for — lab bhattacharjee, Mar 17 '14 at 13:05

score 2 · Answer 1 · answered Mar 17 '14 at 12:50

Apply Cauchy-Schwarz Inequality to argue that,

$(a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$,

And further, $\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab}=\frac{1}{ab}\ge \frac{4}{(a+b)^2}=4$.

So the first expression becomes,

$25=(1+4)^2 \le (a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$

That is $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2 \ge 25/2$.

and Equality holds iff $a=b=\frac12$.

DonAntonio · Answer 2 · 2014-03-17T16:07:54.237

1

Hint:

Put $\;b=1-a\;$ in the left side of the inequality to prove, so:

$$\left(a+\frac1a\right)^2+\left(1-a+\frac1{1-a}\right)^2=a^2+2+\frac1{a^2}+(1-a)^2+2+\frac1{(1-a)^2}=$$

$$=2a^2-2a+5+\frac{2a^2-2a+1}{\left(a(1-a)\right)^2}$$

But it is given that

$$\frac14>ab=a(1-a)=a-a^2\iff a(a-1)=a^2-a>-\frac14\;\ldots$$

Try to take it from here now...and check the inequality sign.

edited Mar 17 '14 at 16:07

answered Mar 17 '14 at 12:42

DonAntonio

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isn't it < in stead of > at the last 1/4... – Ignace Mar 17 '14 at 16:05
No, but in fact it should be $$ab=a(1-a)\le\frac14\implies a(a-1)\ge-\frac14\ldots$$ – DonAntonio Mar 17 '14 at 16:07
given the equation to be solved that gives: x > 1/2 + 5 x meaning 2 times a^2 -a. so 5 + x. x > 1/2 + 1 as nominator. We have to multiply the denominator by -1. – Ignace Mar 17 '14 at 16:22
@Ignace I can't understand your mathematics writing. Please do use LaTeX – DonAntonio Mar 17 '14 at 16:33

score 1 · Answer 3 · answered Mar 18 '14 at 09:47

$$ \begin{align} \hspace{-1cm}\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2 &\ge2\left(a+\frac1a\right)\left(b+\frac1b\right)\tag{1}\\ &=\frac2{ab}\left(a^2+1\right)\left(b^2+1\right)\\ &\ge8\left(a^2+1\right)\left(b^2+1\right)\tag{2}\\ &=8\small\left(\left(a-\tfrac12\right)^2+\left(a-\tfrac12\right)+\tfrac54\right)\left(\left(a-\tfrac12\right)^2-\left(a-\tfrac12\right)+\tfrac54\right)\\ &=8\left[\left(\left(a-\tfrac12\right)^2+\tfrac54\right)^2-\left(a-\tfrac12\right)^2\right]\\ &=8\left[\left(a-\tfrac12\right)^4+\tfrac32\left(a-\tfrac12\right)^2+\tfrac{25}{16}\right]\\[4pt] &\ge\frac{25}{2}\tag{3} \end{align} $$ Inequalities:
$(1)$: $(x-y)^2\ge0\implies x^2+y^2\ge2xy$
$(2)$: $ab\le\tfrac14$
$(3)$: $x^2\ge0$

Solving an equality in 2 variables

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