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Prove that If $a,b,c \in R^+$ Then

$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \gt 2$$

My try:

I assumed:

$$x=\sqrt{\frac{a}{b+c}}$$

$$y=\sqrt{\frac{b}{a+c}}$$

$$x=\sqrt{\frac{c}{b+a}}$$

Then we have:

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)$$

Now by AM GM Inequality we have:

$$\left(\frac{a}{b}+\frac{b}{a}\right) \ge 2$$

So we get:

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \ge 6 \tag{1}$$

Using $(1)$ we need to prove:

$$x+y+z \gt 2$$

Any idea here?

Umesh shankar
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1 Answers1

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Writing cyclic sums for short, $$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\\ \geq\sum_{cyc}\frac{2a}{a+b+c}=2 $$ where the AM-GM inequality was used for each denominator, i.e. $2\sqrt{a(b+c)} \le a + (b+c)$

As Martin R. pointed out, math.stackexchange.com/a/2237066/42969 has the same reasoning, so credits belong to Michael Rozenberg. Here, some extra explanation is given.

Andreas
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  • So is the inequality $\ge 2$? – Umesh shankar Jan 31 '19 at 09:30
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    Yes. To be more precise, equality would only hold in the AM-GM terms if $a = b+c$, $b = a+c$, $c = b+a$ which would entail $a=b=c=0$ which is not possible. So we can exclude this case, and $>$ holds. – Andreas Jan 31 '19 at 09:33
  • So you mean to say when AM GM is used there are some exceptions like this problems, where equality cannot occur. – Umesh shankar Jan 31 '19 at 09:39
  • Well, what do you mean by "exceptions"? I was explaning that equality in AM-GM - as used here - holds if and only if the two terms are equal (in all three cases where AM-GM was used). Then I argued that if $a,b,c \in R^+$, this equality condition cannot occur. So I concluded that indeed $>$ holds. – Andreas Jan 31 '19 at 09:45
  • Did you notice that I already pointed out a possible duplicate target? Your solution was already presented in https://math.stackexchange.com/a/2237066/42969. – Martin R Jan 31 '19 at 09:48
  • @MartinR Martin, thanks for driving my attention to it, indeed I hadn't noticed your comment. I looked at the source you gave: yes it has the same reasoning with a little less explanation. I put this in the text. What are the rules: should I remove my answer? – Andreas Jan 31 '19 at 09:58
  • @Andreas I am actually not clear. I just wanted to know for which values of $a,b,c$ we get $f(a,b,c)=2$. According to answer given by Rozenberg in linked post when $a=b=1$ and $c \to 0$ we get $f(a,b,c)\to 2$.Can you please clarify on this? – Umesh shankar Jan 31 '19 at 10:03
  • Yes, when any two variables are equal (not necessarily 1, just equal) and the third variable $\to 0$, $f\to 2$. Again, $f=2$ occurs if the third variable equals 0 which is ruled out by the condition $a,b,c \in R^+$. – Andreas Jan 31 '19 at 10:08
  • But what i did is: i solved for $c$ when $f(a,b,c)=2$ and $a=b=1$, that is: $$\frac{2}{\sqrt{1+c}}+\sqrt{\frac{c}{2}}=2$$, then real values i am getting are $c=0$ or $c=12.87$. So how come we say $c \to 0$ – Umesh shankar Jan 31 '19 at 10:16
  • $c = 12.87$ does not give the solution 2. – Andreas Jan 31 '19 at 10:24
  • Yes so can we say that $f(a,b,c)$ has no minimum ,but infimum is $2$ – Umesh shankar Jan 31 '19 at 10:27
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    Yes that's how I would put it. – Andreas Jan 31 '19 at 10:31