I'm reading Pham Kim Hung's 'Secrets in Inequalities - Volume 1', and I have to say from the first few examples, that it is not a very good book. Definitely not beginner friendly.

Anyway, it is proven by the author, that for four variables $a, b, c$, and $d$, each being a non-negative real number, the following inequality holds: $$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\ge 2$$

I have no idea how the author proves this. It comes under the very first section, AM-GM. I get the original Nesbitt's inequality in 3 variables that the author proves (which is also cryptic, but I was able to decipher it).

My effort: I understood how the author defines the variables $M, N$ and $S$. $$S = \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$ $$M = \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} + \frac{a}{a+b}$$ $$N = \frac{c}{b+c} + \frac{d}{c+d} + \frac{a}{d+a} + \frac{b}{a+b}$$

$M + N = 4$, pretty straightforward. The numerators and denominators cross out to give four 1s.

Then the author, without any expansion/explanation, says

$$M + S = \frac{a+b}{b+c} + \frac{b+c}{c+d} + \frac{c+d}{d+a} + \frac{d+a}{a+b}\ge 4$$

Which is also true, since the AM-GM inequality says

$$\frac{M+S}{4}\ge \left(\frac{a+b}{b+c}\cdot\frac{b+c}{c+d}\cdot\frac{c+d}{d+a}\cdot\frac{d+a}{a+b}\right)^{1/4}$$

The RHS above evaluates to $1^{1/4}$ since all the numerators and denominators cancel out.

The next part is the crux of my question.

The author claims,

$$N + S =\frac{a+c}{b+c}+\frac{a+c}{a+d}+\frac{b+d}{c+d} + \frac{b+d}{a+b}\ge\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}$$

This is completely bizarre for me! Where did the author manage to get a sum of $(a+b+c+d)$??

As a side note, I'd definitely not recommend this book for any beginner in basic algebraic inequalities (even though the title of the book promotes that it's a treatment of basic inequalities). The author takes certain 'leaps of faith', just assuming that the student reading the book would be able to follow.

  • $\begingroup$ One thing I can understand is (a+c)/(b+c) >= (a+c)/(a+b+c+d) since all four numbers are non-negative. So, (a+c)/(b+c) + (a+c)/(a+d) >= 2(a+c)/(a+b+c+d). Where did the factor of 4 come from? $\endgroup$ Jun 21, 2015 at 20:52
  • $\begingroup$ @mathlove: Thanks for the edit! $\endgroup$ Jun 22, 2015 at 16:47

4 Answers 4


Since we have $(x-y)^2\ge 0$, we have, for $x\gt 0,y\gt 0$, $$\begin{align}(x-y)^2\ge 0&\Rightarrow x^2+y^2+2xy\ge 4xy\\&\Rightarrow y(x+y)+x(x+y)\ge 4xy\\&\Rightarrow \frac{1}{x}+\frac 1y\ge\frac{4}{x+y}\end{align}$$ Now set $x=b+c,y=a+d$ and $x=c+d,y=a+b$ to get $$\frac{1}{b+c}+\frac{1}{a+d}\ge\frac{4}{b+c+a+d}$$and$$\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{c+d+a+b}.$$

  • 1
    $\begingroup$ Thank you! This makes perfect sense now. I completely dislike how the author stepped over, like 5-7 steps you mentioned above, just ASSUMING someone reading the book would understand. $\endgroup$ Jun 21, 2015 at 21:17

The simplest is to use the AM-HM inequality; which is a consequence of the AM-GM inequality, but can be proved independently.

Remember the harmonic mean of $x$ and $y$ is the number whose inverse is the arithmetic mean of the inverses of $x$ and $y$. Explicitly: $$\frac1H=\frac12\Bigl(\frac1x+\frac1y\Bigr)\ge \frac1A=\frac2{x+y}$$ Thus $\,\dfrac1x+\dfrac1y\ge\dfrac4{x+y}$.

Apply this inequality twice in $$(a+c)\Bigl(\frac1{b+c}+\frac1{a+d}\Bigl)+(b+d)\Bigl(\frac1{c+d}+\frac1{a+b}\Bigl). $$


You can indeed use AM-GM, as the book says, to derive the result. Admittedly, it would have been more transparent had the book added one line of derivation as shown below.

For positive $x$ and $y$, $$\frac{x+y}2\,\frac12\Big(\frac1x+\frac1y\Big)\ge\sqrt{xy}\sqrt{\frac1{xy}}=1,$$ thus $$\frac1x+\frac1y\ge \frac4{x+y}.$$ Substituing $x=b+c$ and $y=a+d$, you get the desired inequality.


I think, the best way it's C-S: $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{\left(\sum\limits_{cyc}a\right)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a-c)^2+(b-d)^2}{\sum\limits_{cyc}(ab+ac)}\geq2.$$

Another way:

Let $(a-c)(b-d)\geq0.$

Thus, \begin{align} \sum_{cyc}\frac{a}{b+c}&=\frac{a+d}{b+c}+\frac{b+c}{a+d}+\frac{d}{a+b}-\frac{d}{b+c}+\frac{b}{c+d}-\frac{b}{a+d} \\ &\geq 2+\frac{d(c-a)}{(a+b)(b+c)}+\frac{b(a-c)}{(c+d)(a+d)} \\ &=2+\frac{(a-c)(b-d)(b^2+d^2+ab+ac+ad+bc+bd+cd)}{(a+b)(b+c)(c+d)(d+a)} \\ &\geq2. \end{align}

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    $\begingroup$ Can down-voter explain us, why did you do it? $\endgroup$ Nov 1, 2019 at 21:21
  • $\begingroup$ +1. Can this inequality and proof be generalized to $n$ variables? $\endgroup$
    – Hans
    Nov 2, 2019 at 0:16
  • $\begingroup$ @Hans No. For $n=14$ the inequality is wrong. $\endgroup$ Nov 2, 2019 at 4:01
  • $\begingroup$ Can you give a counterexample? You are saying it is wrong for the lower bound of $\frac n2$, right? Is there a positive lower bound at all? $\endgroup$
    – Hans
    Nov 2, 2019 at 4:05
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    $\begingroup$ @Hans There is a paper in Russian only. Maybe the following: en.wikipedia.org/wiki/Shapiro_inequality $\endgroup$ Nov 2, 2019 at 4:16

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