The equation $(x+a) (x+b) = 9$ has a root $a+b$. Prove that $ab\le 1$.
Problem
Source: Caucasus 2015 11.2
Tags: algebra, inequalities, trinomial
26.04.2019 16:51
parmenides51 wrote: The equation $(x+a) (x+b) = 9$ has a root $a+b$. Prove that $ab\le 1$. Let $a,b $ be reals such that $(2a+b) (2b+a) = 9$ . Prove that $$ab\le 1$$Proof : $$9ab\leq 2(a^2+b^2)+5ab\leq 9.$$
22.11.2021 23:25
(x+a)(x+b)=x^2+x(a+b)+ab x=a+b:2(a+b)^2+ab=9>=2(4ab)+ab=9ab 9ab<=9 ==> ab<=1
22.11.2021 23:30
Mahan2006 wrote: (x+a)(x+b)=x^2+x(a+b)+ab x=a+b:2(a+b)^2+ab=9>=2(4ab)+ab=9ab 9ab<=9 ==> ab<=1
23.11.2021 02:59
Let $a,b $ be reals such that $(a+b) (a+4b) = 9 $ . Prove that $$ab\le 1$$Let $a,b $ be reals such that $(a+b) (a+2b) (a+4b) = 9$ . Prove that $$ab\le \frac12$$
23.11.2021 03:18
sqing wrote: Let $a,b $ be reals such that $(a+b) (a+4b) = 9 $ . Prove that $$ab\le 1$$ It suffices to show: $$9ab\le(a+b)(a+4b)$$$$\Leftrightarrow(a-2b)^2\ge0$$ sqing wrote: Let $a,b $ be reals such that $(a+b) (a+2b) (a+4b) = 9$ . Prove that $$ab\le \frac12$$ Multiplying the AM-GM inequalities: $$\frac12a+\frac12a+b\ge\frac{3\sqrt[3]2}2\sqrt[3]{ab}$$$$a+2b\ge2\sqrt2\sqrt{ab}$$$$a+2b+2b\ge3\sqrt[3]4\sqrt[3]{ab}$$
23.11.2021 03:20
$a,b $ be reals.
23.11.2021 03:48
Let $a,b $ be reals such that $ (2a+b^2) (2b+a^2) =9 $ . Prove that $$ab\le 1$$Let $a,b $ be reals such that $ (a+b^2) (a^2+4b) =9 $ . Prove that $$ab\le 1$$
23.11.2021 04:01
sqing wrote: Let $a,b $ be reals such that $ (2a+b^2) (2b+a^2) =9 $ . Prove that $$ab\le 1$$am-gm $(a+b^2+a)>=3\sqrt[3]{a^2b^2}$ $(b+a^2+b)>=3\sqrt[3]{a^2b^2}$ $9\sqrt[3]{a^4b^4}<=9$ $ab<=1$
23.11.2021 04:03
sqing wrote: Let $a,b $ be reals such that $ (2a+b^2) (2b+a^2) =9 $ . Prove that $$ab\le 1$$Let $a,b $ be reals such that $ (a+b^2) (a^2+4b) =9 $ . Prove that $$ab\le 1$$
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23.11.2021 11:26
If a,b is positive: By Cauchy-Schwarz inequality, 9=(a+b^2)(4b+a^2)>=(2sqrt(ab)+ab)^2 so ab<=1.
23.11.2021 11:48
StarLex1 wrote: sqing wrote: Let $a,b $ be reals such that $ (2a+b^2) (2b+a^2) =9 $ . Prove that $$ab\le 1$$am-gm $(a+b^2+a)>=3\sqrt[3]{a^2b^2}$ $(b+a^2+b)>=3\sqrt[3]{a^2b^2}$ $9\sqrt[3]{a^4b^4}<=9$ $ab<=1$ you only proved for $a,b>0$
23.11.2021 12:48
https://artofproblemsolving.com/community/c6h2724492_inspire_by_caucasus_2015 you can see my prove here it is basically same
23.11.2021 15:57