Define $ \phi$ the positive real root of $ x^2 - x - 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a + 2b)^2 = 4c^2 + 1$. Show that $ \displaystyle 2d^2 + a^2\left(\phi - \frac {1}{2}\right) + b^2\left(\frac {1}{\phi - 1} + 2\right) + 2 \ge 4(c - d) + 2\sqrt {d^2 + 2d}$ and find all cases of equality. (A.Naskov)
Problem
Source: Oliforum math contest, problem 2
Tags: inequalities, inequalities proposed
Bugi
30.09.2009 23:25
Let $ x=\sqrt{d^2+2d}$
Then $ 2d^2+4d-2\sqrt{d^2+2d}=2x^2-2x\ge -0.5$ by AM-GM
So if $ y=a^2(\phi-\frac12)+b^2(\frac1{\phi-1}+2)+2-4c\ge 0.5$, we have proved the inequality.
Expand and rearrange
$ a^2(2\phi-1)+b^2(\frac2{\phi-1}+4)\ge 8c-3$
We know that $ \phi^2=\phi+1$, and $ \frac1{\phi}=\phi-1$
Now $ 2\phi-1=1+\frac{2}{\phi}$, and $ \frac2{\phi-1}+4=2\phi+4$
We need to prove $ (a^2+4b^2)+(\frac{2a^2}{\phi}+2b^2\phi)\ge 8c-3$
$ a^2+4b^2=(a+2b)^2-4ab=4c^2+1-4ab$
$ \frac{2a^2}{\phi}+2b^2\phi-4ab\ge -4(c-1)^2$
LHS is non-negative by AM-GM, and inequality is proved
Equality case: $ d=1-d$, so $ d=0.5$
$ c-1=0$, so $ c=1$
$ a=b\phi,(a+2b)^2=5$, so $ a=\frac{\phi}{\sqrt{\phi+1}},b=\frac1{\sqrt{\phi+1}}$
Zhero
01.10.2009 00:18
Woah that was intense. Just as a note, you could simply them further to get $ a = 1$ and $ b = \frac{\sqrt{5} - 1}{2}$. Also, I got $ d = \frac{\sqrt{5} - 2}{2}$ for the equality case for $ d$...?
Bugi
01.10.2009 00:21
Oh fail. But what about d? Is there a mistake in my solution? Can you show yours?
can_hang2007
01.10.2009 06:24
bboypa wrote: Define $ \phi$ the positive real root of $ x^2 - x - 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a + 2b)^2 = 4c^2 + 1$. Show that $ \displaystyle 2d^2 + a^2\left(\phi - \frac {1}{2}\right) + b^2\left(\frac {1}{\phi - 1} + 2\right) + 2 \ge 4(c - d) + 2\sqrt {d^2 + 2d}$ and find all cases of equality. (A.Naskov) Notice first that $ \phi >1.$ Applying the Cauchy Schwarz Inequality, we have \[ \left[ a^{2}\left( \phi -\frac{1}{2}\right) +b^{2}\left( \frac{1}{\phi -1}+2\right) \right] \left( \dfrac{1}{\phi -\dfrac{1}{2}}+\dfrac{4}{\dfrac{1}{\phi -1}+2}\right) \geq (a+2b)^{2}.\] On the other hand, it is easy to verify that \[ \dfrac{1}{\phi -\dfrac{1}{2}}+\dfrac{4}{\dfrac{1}{\phi -1}+2}=2.\] Combining this with the above inequality, we get \[ a^{2}\left( \phi -\frac{1}{2}\right) +b^{2}\left( \frac{1}{\phi -1}+2\right) \geq \frac{(a+2b)^{2}}{2}=\frac{4c^{2}+1}{2}.\] Thus, it is enough to prove that \[ 2d^{2}+2c^{2}+\frac{5}{2}\geq 4(c-d)+2\sqrt{d^{2}+2d}.\] This inequality is equivalent to \[ 2(c-1)^{2}+\frac{1}{2}\left( 2\sqrt{d^{2}+2d}-1\right) ^{2}\geq 0,\] which is obviously true.
Bugi
01.10.2009 10:42
Zhero, you are right, I made a mistake for d. It's in fact $ x=1-x$, so $ x=0.5$, but then $ d$ is what you said...
bboypa
07.10.2009 08:42
bboypa wrote: Define $ \phi$ the positive real root of $ x^2 - x - 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a + 2b)^2 = 4c^2 + 1$. Show that $ \displaystyle 2d^2 + a^2\left(\phi - \frac {1}{2}\right) + b^2\left(\frac {1}{\phi - 1} + 2\right) + 2 \ge 4(c - d) + 2\sqrt {d^2 + 2d}$ and find all cases of equality. (A.Naskov) Define $ e: =d-1,f: =2b$, then only by am-gm: $ \displaystyle 2e^2+a^2\left(\phi - \frac {1}{2}\right) + f^2\left(\frac {1}{4(\phi - 1)} + \frac{1}{2}\right)-4c=$ $ \displaystyle \frac{\sqrt{5}(\sqrt{5}-1)}{2}\left(\frac{4e^2}{5}+\frac{f^2}{(\sqrt{5}-1)^2} \right) + \frac{a^2\sqrt{5}}{2} + \frac{2e^2}{\sqrt{5}} -4c$ $ \displaystyle \ge 2ae+2fe-4c=$ $ (e+1)(a+f-2c)+(e-1)(a+f+2c) \ge 2\sqrt{(e^2-1)((a+f)^2-4c^2)}$ $ =2\sqrt{e^2-1}$, that is our aim.[]