If real numbers $a>b>1$ satisfy the inequality$$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$$what is the minimum possible value of $\dfrac{\sqrt{a-b}}{b-1}$
Problem
Source: 2017 Turkey Junior National Olympiad
Tags: inequalities, factorization
CinarArslan
17.12.2017 18:25
$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$
$\implies 0\leq 2a^3+2b^3-a^2b^2-a^2-b^2-4ab+2a+2b-1$
$\implies 0\leq (2a^3-4ab+2a)+(-a^2b^2+2b^3-b^2)+(-a^2+2b-1)$
$\implies 0\leq (-a^2+2b-1)(-2a)+(-a^2+2b-1)(b^2)+(-a^2+2b-1)(1)$
$\implies 0\leq (-a^2+2b-1)(b^2-2a+1)$
$\implies 0\leq (a^2-2b+1)(2a-b^2-1)$
And $a>b>1 \implies a^2>b^2 \implies a^2-2b+1>b^2-2b+1=(b-1)^2>0$
$\implies a^2-2b+1>0$
So $2a-b^2-1\geq 0$
$\implies 2a\geq b^2+1 \implies 2a-2b\geq b^2-2b+1$ , $b>1 \implies \sqrt {2(a-b)}\geq b-1$
$\implies \frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$
This inequality holds when $2a=b^2+1$. Examples: $(a,b)={{(\frac{5}{2},2),(5,3)(\frac{17}{2},4)...}}$
CinarArslan
17.12.2017 18:26
But it was not easy to see that $2a^3+2b^3-a^2b^2-a^2-b^2-4ab+2a+2b-1=(a^2-2b+1)(2a-b^2-1)$
sqing
18.12.2017 04:04
CinarArslan wrote:
$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$
$\implies 0\leq 2a^3+2b^3-a^2b^2-a^2-b^2-4ab+2a+2b-1$
$\implies 0\leq (2a^3-4ab+2a)+(-a^2b^2+2b^3-b^2)+(-a^2+2b-1)$
$\implies 0\leq (-a^2+2b-1)(-2a)+(-a^2+2b-1)(b^2)+(-a^2+2b-1)(1)$
$\implies 0\leq (-a^2+2b-1)(b^2-2a+1)$
$\implies 0\leq (a^2-2b+1)(2a-b^2-1)$
And $a>b>1 \implies a^2>b^2 \implies a^2-2b+1>b^2-2b+1=(b-1)^2>0$
$\implies a^2-2b+1>0$
So $2a-b^2-1\geq 0$
$\implies 2a\geq b^2+1 \implies 2a-2b\geq b^2-2b+1$ , $b>1 \implies \sqrt {2(a-b)}\geq b-1$
$\implies \frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$
When $a=\frac{5}{2},b=2$ , have $ \frac {\sqrt {a-b}}{b-1}=\frac{1}{\sqrt{2}}.$ If $a>b>1$ and $(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$ . Then$$\frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$$
CinarArslan
18.12.2017 11:20
mistake...
AnArtist
18.12.2017 11:23
@above he is just completing your proof to show that equality holds.
CinarArslan
18.12.2017 11:25
AnArtist wrote: @above he is just completing your proof to show that equality holds. Oh okay, thank you
sqing
18.12.2017 11:28
AnArtist wrote: @above he is just completing your proof to show that equality holds. Yeah. Thanks.
CinarArslan
18.12.2017 11:32
sqing wrote: CinarArslan wrote:
$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$
$\implies 0\leq 2a^3+2b^3-a^2b^2-a^2-b^2-4ab+2a+2b-1$
$\implies 0\leq (2a^3-4ab+2a)+(-a^2b^2+2b^3-b^2)+(-a^2+2b-1)$
$\implies 0\leq (-a^2+2b-1)(-2a)+(-a^2+2b-1)(b^2)+(-a^2+2b-1)(1)$
$\implies 0\leq (-a^2+2b-1)(b^2-2a+1)$
$\implies 0\leq (a^2-2b+1)(2a-b^2-1)$
And $a>b>1 \implies a^2>b^2 \implies a^2-2b+1>b^2-2b+1=(b-1)^2>0$
$\implies a^2-2b+1>0$
So $2a-b^2-1\geq 0$
$\implies 2a\geq b^2+1 \implies 2a-2b\geq b^2-2b+1$ , $b>1 \implies \sqrt {2(a-b)}\geq b-1$
$\implies \frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$
When $a=\frac{5}{2},b=2$ , have $ \frac {\sqrt {a-b}}{b-1}=\frac{1}{\sqrt{2}}.$ If $a>b>1$ and $(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$ . Then$$\frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$$ Thank you sqing I edited the proof
sqing
18.12.2017 11:39
CinarArslan wrote: sqing wrote: CinarArslan wrote:
$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$
$\implies 0\leq 2a^3+2b^3-a^2b^2-a^2-b^2-4ab+2a+2b-1$
$\implies 0\leq (2a^3-4ab+2a)+(-a^2b^2+2b^3-b^2)+(-a^2+2b-1)$
$\implies 0\leq (-a^2+2b-1)(-2a)+(-a^2+2b-1)(b^2)+(-a^2+2b-1)(1)$
$\implies 0\leq (-a^2+2b-1)(b^2-2a+1)$
$\implies 0\leq (a^2-2b+1)(2a-b^2-1)$
And $a>b>1 \implies a^2>b^2 \implies a^2-2b+1>b^2-2b+1=(b-1)^2>0$
$\implies a^2-2b+1>0$
So $2a-b^2-1\geq 0$
$\implies 2a\geq b^2+1 \implies 2a-2b\geq b^2-2b+1$ , $b>1 \implies \sqrt {2(a-b)}\geq b-1$
$\implies \frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$
When $a=\frac{5}{2},b=2$ , have $ \frac {\sqrt {a-b}}{b-1}=\frac{1}{\sqrt{2}}.$ If $a>b>1$ and $(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$ . Then$$\frac {\sqrt {a-b}}{b-1}\geq \frac{1}{\sqrt{2}}$$ Thank you sqing I edited the proof Your solution is very exciting. Thanks.
CinarArslan
18.12.2017 12:15
sqing wrote: Your solution is very exciting. Thanks. Thanks
RagvaloD
08.08.2018 13:30
$$(ab+1)^2+(a+b)^2\leq 2(a+b)(a^2-ab+b^2+1)$$$(ab+1)^2+(a+b)^2-2(ab+1)(a+b)\leq 2(a+b)(a^2-ab+b^2+1)-2(ab+1)(a+b)$ $(a-1)^2(b-1)^2 \leq 2(a+b)(a-b)^2$ Let $b=x+1,a=b+kx^2=kx^2+x+1,k>0$ $(kx^2+x)^2 x^2 \leq 2(kx^2+2x+2)k^2x^4$ and we need to find minimum of $\sqrt{k}$ $(kx+1)^2 \leq 2(kx^2+2x+2)k^2$ $(kx+1)^2 \leq 2k(kx+1)^2+2k(2k-1)$ $(1-2k) ((kx+1)^2+2k) \leq 0$ $2k \geq 1$
Ilove_mathematics
25.10.2018 18:02
Omg i solved this problem in 30 minutes, the same way from CinarArslan. The hard part was to see the factorization.