MHB Prove $40ac<1$ Given $a, b, c$ Real Positive Numbers

[anemone]

Let $a,\,b,\,c$ be three distinct positive real numbers such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}$, prove that $40ac<1$.

 

[Albert1]

Let $a,\,b,\,c$ be three distinct positive real numbers such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}$, prove that $40ac<1$.

must have a typo ,I think it should be:
prove that: $ac<1$
check :$b=3,c=1$ and we get $0.5<a<1$
if $(a\neq c=1)$

 

[anemone]

must have a typo ,I think it should be:
prove that: $ac<1$
check :$b=3,c=1$ and we get $0.5<a<1$
if $(a\neq c=1)$

Hi Albert,

I just checked with my source and no, there is no typo and although I can't provide an example to show that $40ac<1$ is true under the given condition, I suppose the problem is correct, judging from the solution that I have at hand...

But, if $b=3$, I then checked with wolfram (solve x'+'sqrt'{'3'+''\'sqrt'{'y'}''}''='y'+'sqrt'{'3'+''\'sqrt'{'x'}''}' - Wolfram|Alpha) and I hoped to obtain at least one set of solution for $(a,\,3,\,c)$ to show that $40ac<1$ is true, I realized the second relationship between $a$ and $c$ is what I wanted, and by plugging $a=1$, it returned a negative $c$, that has thrown me off. $c$ can't be a negative...

Perhaps $b$ lies in certain interval and it certainly can't be a 3...

 

[Albert1]

Hi Albert,

I just checked with my source and no, there is no typo and although I can't provide an example to show that $40ac<1$ is true under the given condition, I suppose the problem is correct, judging from the solution that I have at hand...

But, if $b=3$, I then checked with wolfram (solve x'+'sqrt'{'3'+''\'sqrt'{'y'}''}''='y'+'sqrt'{'3'+''\'sqrt'{'x'}''}' - Wolfram|Alpha) and I hoped to obtain at least one set of solution for $(a,\,3,\,c)$ to show that $40ac<1$ is true, I realized the second relationship between $a$ and $c$ is what I wanted, and by plugging $a=1$, it returned a negative $c$, that has thrown me off. $c$ can't be a negative...

Perhaps $b$ lies in certain interval and it certainly can't be a 3...

Perhaps $b$ lies in certain interval and it certainly can't be a 3---
To this problem it only said $(a\neq b \neq c)$, and $(a,b,c>0)$
if $b$ can't be 3 ,you should point out the interval of $ a,b,c$ first

 

[Nono713]

Perhaps $b$ lies in certain interval and it certainly can't be a 3---
To this problem it only said $(a\neq b \neq c)$, and $(a,b,c>0)$
if $b$ can't be 3 ,you should point out the interval of $ a,b,c$ first

I'm not seeing the issue here. The problem clearly asks to prove that $40ac < 1$ whenever $a + \sqrt{b + \sqrt{c}} = c + \sqrt{b + \sqrt{a}}$ holds for distinct reals $a, b, c > 0$. That equation is what gives the necessary restrictions on $a$, $b$ and $c$, as not all triplets of positive reals $(a, b, c)$ satisfy the equation. If $b$ can't be 3 for the equation to hold then that just means that there is nothing to prove for triplets of the form $(a, 3, c)$, and the fact that $40ac < 1$ fails to hold for these particular triplets is immaterial.

 

[Opalg]

Let $a,\,b,\,c$ be three distinct positive real numbers such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}$, prove that $40ac<1$.

[sp]The problem is symmetric in $a$ and $c$, so we may assume that $a>c$. Then $a-c = \sqrt{b+\sqrt a} - \sqrt{b+\sqrt c}$ and so $$ \begin{aligned} (a-c)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) &= \bigl(\sqrt{b+\sqrt a} - \sqrt{b+\sqrt c}\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) \\ &= \bigl(b+\sqrt a\bigr) - \bigl(b+\sqrt c\bigr) = \sqrt a - \sqrt c. \end{aligned}$$ Therefore $(a-c)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = \bigl(\sqrt a - \sqrt c\bigr)\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = \sqrt a - \sqrt c$. Since $a\ne c$, we can divide through by $\sqrt a - \sqrt c$ to get $$\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = 1.$$ For that to have a solution with $b>0$ we must have $\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt[4]{ a} + \sqrt[4]{c}\bigr) < 1.$ Thus $$\tfrac12\bigl(\sqrt a + \sqrt c\bigr) \cdot\tfrac12\bigl(\sqrt[4]{ a} + \sqrt[4]{c}\bigr) <\tfrac14.$$ By GM-AM, that implies $\sqrt[4]{ac}\,\sqrt[8]{ac} < \frac14$, so that $(ac)^{3/8} < \frac14$ and hence $4^{8/3}ac < 1$. But $4^{8/3} \approx 40.3$, so we conclude that $40ac<1.$[/sp]

 

[Albert1]

Let $a,\,b,\,c$ be three distinct positive real numbers such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}----(1)$, prove that $40ac<1$.

putting $b=3,c=1$ to (1) and simplify , I get :
$a^4+4a^3-9a+4=0=(a-1)(a^3+5a^2+5a-4)$
since $a \neq 1 ,\therefore a^3+5a^2+5a-4=0----(2)$
the only solution for (2) is $a\approx 0.5115$
this leads to $ac<1$
how to explain it?

 

[Opalg]

putting $b=3,c=1$ to (1) and simplify , I get :
$a^4+4a^3-9a+4=0=(a-1)(a^3+5a^2+5a-4)$
since $a \neq 1 ,\therefore a^3+5a^2+5a-4=0----(2)$
the only solution for (2) is $a\approx 0.5115$
this leads to $ac<1$
how to explain it?

If $a = 0.5115$, $b=3$ and $c=1$ then $a + \sqrt{b + \sqrt c} = 0.5115 + \sqrt4 = 2.5115$. But $c + \sqrt{b + \sqrt a} \approx 1 + \sqrt{3 + 0.7152} \approx 1 + 1.9275 = 2.9275 \ne 2.5115.$ So the value found for $a$ is a spurious result, not a solution of the original equation. It may have been introduced by squaring at some stage.

 

[anemone]

[sp]The problem is symmetric in $a$ and $c$, so we may assume that $a>c$. Then $a-c = \sqrt{b+\sqrt a} - \sqrt{b+\sqrt c}$ and so $$ \begin{aligned} (a-c)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) &= \bigl(\sqrt{b+\sqrt a} - \sqrt{b+\sqrt c}\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) \\ &= \bigl(b+\sqrt a\bigr) - \bigl(b+\sqrt c\bigr) = \sqrt a - \sqrt c. \end{aligned}$$ Therefore $(a-c)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = \bigl(\sqrt a - \sqrt c\bigr)\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = \sqrt a - \sqrt c$. Since $a\ne c$, we can divide through by $\sqrt a - \sqrt c$ to get $$\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt{b+\sqrt a} + \sqrt{b+\sqrt c}\bigr) = 1.$$ For that to have a solution with $b>0$ we must have $\bigl(\sqrt a + \sqrt c\bigr)\bigl(\sqrt[4]{ a} + \sqrt[4]{c}\bigr) < 1.$ Thus $$\tfrac12\bigl(\sqrt a + \sqrt c\bigr) \cdot\tfrac12\bigl(\sqrt[4]{ a} + \sqrt[4]{c}\bigr) <\tfrac14.$$ By GM-AM, that implies $\sqrt[4]{ac}\,\sqrt[8]{ac} < \frac14$, so that $(ac)^{3/8} < \frac14$ and hence $4^{8/3}ac < 1$. But $4^{8/3} \approx 40.3$, so we conclude that $40ac<1.$[/sp]

Thanks Opalg,:) for participating in this challenge and thanks too for the explanation that answered to Albert's question.