To prove the given inequality

Question

Question:-

If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$

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Attempt at a solution:-

I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-

$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge \sqrt{\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b}}$$

Evaluating the right hand side of the inequality,

$$\left(\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b} \right) = \left( \dfrac{b^2+\left(a+c \right)b+ac}{b^2-2b\left(a+c \right)+4ac}\right)$$

Now, as $a,b,c$ are in H.P., hence we get the relation $$b=\dfrac{2ac}{a+c}$$ On substituting this result in the equation, we get

$$\dfrac{10 a^2c^2 + 3ac(a^2+c^2)}{4a^2c^2}=\dfrac{5}{2}+\dfrac{3(a^2+c^2)}{4ac}$$

Now from the above calculations, what I observed is that, that for the proof to be valid, the condition $\dfrac{3(a^2+c^2)}{4ac}=\dfrac{3}{2}$ should be satisfied. The final conclusion to which I arrived was $a=c$. Now although the numbers would still be in H.P. but it doesn't prove the result for the numbers which are in H.P. and are not equal.

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Any other approach to the proof is invited too.

Answer 1

So, why don't we rewrite the HP condition in a nicer way, as $$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$ and let's multiply the numerator and denominator of the first fraction by $2/b$. We get $$ \frac{a+b}{2a-b} = \frac{a(1/a+1/c)+2}{2a(1/a+1/c)-2} = \frac{3+a/c}{2a/c}=\frac{3c+a}{2a}$$ Similarly, the other one becomes $\frac{3a+c}{2c}$, so what we want to show reduces to $$\frac{3c+a}{2a} + \frac{3a+c}{2c}\geq 4$$ which reduces to $$\frac{c}{a}+\frac{a}{c}\geq 2$$ which follows directly from AM-GM.

Answer 2

The theorem you are asked to prove is not true in general. The additional condition needed is that $a \geq b$. And then you can do the proof without AM-GM.

Assuming $(a,b,c)$ are in H.P. in the order $$a = \frac{1}{x}, b = \frac{1}{x+d}, c = \frac{1}{x+2d}$$ then the LHS is $$ \frac{4x^2+8dx+6d^2}{x^2+2dx} $$ Consider the case of $x^2+2dx > 0$: $$ \frac{4x^2+8dx+6d^2}{x^2+2dx}-4 = \frac{1}{x^2+2d}\left[4x^2+8dx+6d^2 - 4(x^2+2dx) \right] = \frac{6d^2}{x^2+2d} \geq =0 $$ Now consider the case where $x^2+2dx < 0$ and $d\neq 0$: $$ \frac{4x^2+8dx+6d^2}{x^2+2dx}-4 = \frac{6d^2}{x^2+2d} < 0 $$ So a good counterexample would be $x=3, d=-1$, giving $a=\frac13, b = \frac12, c=1$.

If you add the condition that $a\geq b$ then you find that forces $x^2 + 2dx \geq 0$ and the statement then is true.