the problem

Let $a,b,c,d\in R^*_+$ so every of the 4 number is smaller than the sum of the other 3. Show that

$\frac{1}{a+b+c}+\frac{1}{a+b+d}+ \frac{1}{a+d+c}+\frac{1}{d+b+c}< \frac{6}{a+b+c+d}$

my idea

So we know that $a< b+c+d, b<a+c+d, c<a+b+d, d<a+b+d$.

I tried using this $ \frac{1}{a+b+c}+ \frac{1}{a+b+d}+ \frac{1}{a+d+c}+ \frac{1}{d+b+c} < \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}$ but I don't know what to do forward

Hope one of you can help me! Thank you!

  • 1
    $\begingroup$ I'm pretty sure that since $a+b+c<a+b+c+d$, $1/(a+b+c)>1/(a+b+c+d)$. $\endgroup$ Commented Jul 10 at 11:57
  • 1
    $\begingroup$ I feel we should replace the word 'so' in the first line with 'in such a way that'. $\endgroup$
    – Saeed
    Commented Jul 10 at 13:57
  • $\begingroup$ @Lucenaposition Yes. From there, it follows that $\frac{1}{a+b+c} + \frac{1}{a+b+d} + \frac{1}{a+c+d} + \frac{1}{b+c+d} > \frac{4}{a+b+c+d}$ which doesn't violate the problem statement. $\endgroup$
    – Saeed
    Commented Jul 10 at 14:00

1 Answer 1


Let's denote $$(x,y,z,t)=\left(\frac{a}{a+b+c+d},\frac{b}{a+b+c+d},\frac{c}{a+b+c+d},\frac{d}{a+b+c+d} \right)$$ Then the problem is equivalent to prove $$\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}+\frac{1}{1-t}<6$$ given $0<x,y,z,t<1/2$ and $x+y+z+t = 1$

We notice that $$ x\left(x-\frac{1}{2} \right)<0 \iff (1-x)(2x+1)>1\iff\color{red}{\frac{1}{1-x}<2x+1} \tag{1}$$ From $(1)$ and $x+y+z+t=1$, we deduce:

$$\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}+\frac{1}{1-t}<(2x+1)+(2y+1)+(2z+1)+(2t+1)=6$$ Q.E.D

  • $\begingroup$ You have a typo in the repeated $1/(1-x)$. $\endgroup$ Commented Jul 10 at 18:36
  • $\begingroup$ @martycohen Thank you! The typo is corrected. $\endgroup$
    – NN2
    Commented Jul 10 at 18:48

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