6
$\begingroup$

I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.

Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$

(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

$\endgroup$
  • 2
    $\begingroup$ Please avoid using $$__$$ in the title. $\endgroup$ – Sasha Aug 31 '12 at 5:19
-3
+50
$\begingroup$

First, I will prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$, and then I will prove that $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}$.

1) $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$

Let's call $\varphi(x_1,x_2,x_3)=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_1}$. Le'ts prove that $\varphi$ is minimized when $x_1 = x_2 = x_3$: $\frac{\partial\varphi}{\partial x_i} = \frac{1}{x_{i+1}} - \frac{x_{i-1}}{x_i^2} = \frac{x_i^2 - x_{i-1}x_{i+1}}{x_i^2 x_{i+1}}$ (with $x_4 = x_1$ and $x_0 = x_3$).

So $\Delta\varphi = 0$ is equivalent to : $$x_1^2 = x_2 x_3$$ $$x_2^2 = x_1 x_3$$ $$x_3^2 = x_1 x_2$$

By dividing one equation by another, we get that $x_1 = x_2 = x_3$. We need to prove that $\varphi$ is convex, to show that this function is minimized when its gradient is $0$. For this, let's compute its Hessian matrix $M$:

$$M = (M_{ij}) = (\frac{\partial^2\varphi}{\partial x_i\partial x_j})$$ We have: $$M_{ii} = \frac{2 x_{i-1}}{x_i^3}$$ $$M_{i(i+1)} = -\frac{1}{x_{i+1}^2}$$ $$M_{i(i-1)} = -\frac{1}{x_i^2}$$

Which gives us:

$$M = \left( \begin{array}{ccc} \frac{2 x_3}{x_1^3} && -\frac{1}{x_2^2} && -\frac{1}{x_1^2}\\ -\frac{1}{x_2^2} && \frac{2 x_1}{x_2^3} && -\frac{1}{x_3^2}\\ -\frac{1}{x_1^2} && -\frac{1}{x_3^2}&& \frac{2 x_2}{x_3^3} \end{array} \right)$$

Calculating its determinant and verifying it is positive will show that $\varphi$ is convex, which achives to prove inequality 1)

2) $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}\geq 1$ This is equivalent to $\sqrt[3]{abc}\geq\frac{a+b+c}{3}$, or $\ln(\frac{a+b+c}{3})\geq \frac{1}{3}(\ln a +\ln b+\ln c)$, which is true because the function $\ln$ is concave.

By combining those two inequalities, we instantly solve the two questions.

$\endgroup$
  • $\begingroup$ in part 2 your inequality is in the wrong direction. By AM-GM arithmetic average is larger than geometric average. $\endgroup$ – Maesumi Sep 12 '12 at 0:18
  • $\begingroup$ @Maesumi that's right, I made a mistake after taking the log, corrected now. Thanks for pointing it out! $\endgroup$ – S4M Sep 12 '12 at 7:54
  • 1
    $\begingroup$ @S4M By AM-GM, $\sqrt[3]{abc}\leq\frac{a+b+c}{3}$. This is a well-known inequality. $\endgroup$ – Arthur Sep 12 '12 at 7:59
  • 4
    $\begingroup$ -1 Part 2 is wrong, as pointed out by Arthur. @Iuli Please unaccept this. $\endgroup$ – Calvin Lin Nov 11 '13 at 14:48
15
$\begingroup$

Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.

You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$

Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.

$\endgroup$
1
$\begingroup$

By C-S $$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$ Thus, it's enough to prove that $$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality it's $f(v^2)\geq0,$ where $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $v^2$, which happens for equality case of two variables.

Since the last inequality is homogeneous, we can assume $b=c=1$. Also, let $a=x^3$.

Id est, we need to prove that $$\frac{(x^3+2)^2}{2x^3+1}+\frac{3x}{x^3+2}\geq4$$ or $$(x-1)^2(x^6+2x^5+3x^4+2x^3+x^2+6x+3)\geq0.$$ Done!

The following inequality is also true.

Let $a$, $b$ and $c$ be positives. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{24\sqrt[3]{abc}}{a+b+c}\geq11.$$

$\endgroup$
  • $\begingroup$ what is C-S? Can you explain $\endgroup$ – user500668 Aug 2 '18 at 15:28
  • $\begingroup$ It's Cauchy-Schwarz inequality: For all reals $a_i$ and positives $b_i$ we have: $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$ $\endgroup$ – Michael Rozenberg Aug 2 '18 at 16:02
  • $\begingroup$ oh thanks, I didn't know that. $\endgroup$ – user500668 Aug 2 '18 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.