Problem related to series both HP and AP

Question

If $a,b,c$ are in H.P. and $\left( {\frac{{a + b}}{{2a - b}}} \right) + \left( {\frac{{c + b}}{{2c - b}}} \right) \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } $ , then find the least value of $\lambda$. (where a, b, c are positive)

My approach is as follow $\frac{1}{a} = a',\frac{1}{b} = b',\frac{1}{c} = c'$ are in A.P.

$2b' = a' + c'$

$\frac{{ab}}{{ab}}\left( {\frac{{\frac{1}{b} + \frac{1}{a}}}{{\frac{2}{b} - \frac{1}{a}}}} \right) + \frac{{bc}}{{bc}}\left( {\frac{{\frac{1}{b} + \frac{1}{c}}}{{\frac{2}{b} - \frac{1}{c}}}} \right) \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } \Rightarrow \left( {\frac{{b' + a'}}{{2b' - a'}}} \right) + \left( {\frac{{b' + c'}}{{2b' - c'}}} \right) \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } $

$\left( {\frac{{b' + a'}}{{c'}}} \right) + \left( {\frac{{b' + c'}}{{a'}}} \right) \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } $

$\frac{{b'}}{{c'}} + \frac{{a'}}{{c'}} + \frac{{b'}}{{a'}} + \frac{{c'}}{{a'}} \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } \Rightarrow \frac{c}{b} + \frac{c}{a} + \frac{a}{b} + \frac{a}{c} \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } \Rightarrow \frac{c}{b} + \frac{c}{a} + \frac{a}{b} + \frac{a}{c} \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } $

$\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } = {\lambda ^{1 + \frac{1}{2} + \frac{1}{4} + .. + \infty }} = {\lambda ^{\frac{1}{{1 - \frac{1}{2}}}}} = {\lambda ^2} \Rightarrow \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } = \sqrt {{\lambda ^2}} = \lambda $

$\frac{{a + c}}{b} + \left( {\frac{c}{a} + \frac{a}{c}} \right) \le \lambda \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \Rightarrow \frac{2}{b} = \frac{{a + c}}{{ac}} \Rightarrow \frac{{a + c}}{b} = \frac{{{{\left( {a + c} \right)}^2}}}{{2ac}}$

$\frac{{{{\left( {a + c} \right)}^2}}}{{2ac}} + \left( {\frac{{2{c^2} + 2{a^2}}}{{2ac}}} \right) \le \lambda $

AFter these steps what I need to do

Answer 1

Right-hand side is equal to $\lambda$.

As $\frac1a + \frac1c = \frac2b$ we have $$\frac{a}b = \frac{a+c}{2c}, \ \frac{c}a = \frac{a+c}{2a}$$

Left-hand side is $$\frac{\frac{a}b + 1}{2\frac{a}b -1} + \frac{\frac{c}b + 1}{2\frac{c}b -1} = \frac{3c+a}{c+2a} + \frac{c+3a}{2c+a} $$ $$=\frac{3t+1}{t+2} + \frac{t+3}{2t+1},$$ where $t = \frac{c}a$. Thus $\lambda = \sup_{t > 0} g(t)$, where $$g(t) = \frac{3t+1}{t+2} + \frac{t+3}{2t+1}.$$ As $g'(t) = \frac{15(t^2-1)}{(t+2)^2(2t+1)^2}$ we have $\sup_{t > 0} g(t) = \max(g(0), g(+\infty) ) = \frac72$. Answer is $\frac72$.

Answer 2

If $a,b,c$ are in H.P. and $\left( {\frac{{a + b}}{{2a - b}}} \right) + \left( {\frac{{c + b}}{{2c - b}}} \right) \le \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } $ , then find the least value of $\lambda$. (where a, b, c are positive)

It was too difficult to move forward on the original poster's work, so I started from scratch.

Suppose $u = \sqrt {\lambda \sqrt {\lambda \sqrt {\lambda ....\infty } } } > 0.$

Then $u^2 = \lambda u \implies u = \lambda.$

So, the question is what is the minimum value for $\lambda$?

Answer : $\lambda = 4$, which is achieved when $a = b = c.$

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Proof

By symmetry of the LHS (which is to be minimized), without loss of generality, $0 < a \leq b \leq c.$

Assume that $b = ka \implies k \geq 1.$

Then
$\frac{1}{a} - \frac{1}{b} = \frac{k-1}{ka} = \frac{k-1}{b}.$

Then
$\frac{1}{b} - \frac{1}{c} = \frac{k-1}{b} \implies $
$\frac{1}{b} - \frac{k-1}{b} = \frac{1}{c} \implies $
$\frac{2 - k}{b} = \frac{1}{c}$.

So, $k < 2.$

Let $f(k) = \left(\frac{a + b}{2a - b}\right) + \left( \frac{c + b}{2c - b} \right)$.
So, $f(k)$ is to be minimized, where $1 \leq k < 2.$

$\displaystyle f(k) = \left(\frac{\frac{b}{k} + b}{\frac{2b}{k} - b}\right) + \left(\frac{\frac{b}{2-k} + b}{\frac{2b}{2-k} - b}\right).$

$\displaystyle f(k) = \left(\frac{\frac{1}{k} + 1}{\frac{2}{k} - 1}\right) + \left(\frac{\frac{1}{2-k} + 1}{\frac{2}{2-k} - 1}\right).$

$\displaystyle f(k) = \left(\frac{1 + k}{2 - k}\right) + \left(\frac{3 - k}{k}\right) = \left(\frac{-2 + k}{2 - k}\right) + \left(\frac{3}{2 - k}\right) + \left(\frac{3}{k}\right) + \left(\frac{- k}{k}\right).$

$\displaystyle f(k) = -1 + 3\left[\frac{1}{2 - k} + \frac{1}{k}\right] - 1.$

Clearly, $f(k)$ may be minimized by minimizing
$\displaystyle \frac{1}{2 - k} + \frac{1}{k} = \frac{2 - k + k}{2k - k^2} = \frac{2}{2k - k^2}.$

Let $g(k) = 2k - k^2 ~: 1 \leq k < 2$.
Then $f(k)$ may be minimized by maximizing $g(k)$.

$g'(k) = 2 - 2k~~~$ and $~~~~g''(k) = -2.$

Therefore, $g(k)$ is maximized at $k = 1$, which minimizes $f(k)$.

Then $k = 1 \implies a = b = c.$