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

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.

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!

• I'm pretty sure that since $a+b+c<a+b+c+d$, $1/(a+b+c)>1/(a+b+c+d)$. Commented Jul 10 at 11:57
• I feel we should replace the word 'so' in the first line with 'in such a way that'. Commented Jul 10 at 13:57
• @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. Commented Jul 10 at 14:00

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 $$0 and $$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
• You have a typo in the repeated $1/(1-x)$. Commented Jul 10 at 18:36