I need to prove:
for a,b,c > 0
prove a/(b+c) + b/(a+c) + c/(b+a) is less than or equal to 3/2
it can be proved by using marvelous AM-GM inequality of positive real numbers.Originally Posted by Nichelle14
let x=a/(b+c) + b/(a+c) + c/(b+a)
now, b+c>=2(bc)^(1/2)
hence,1/(b+c)<=1/2(bc)^(1/2)
hence,2x<=a/(bc)^(1/2) + b/(ac)^(1/2) + c/(ba)^(1/2)
again applying AM-GM inequality on the right hand side of the equation
2x<=3
x<=3/2
AM/GM is your friend. Here's another way, a bit longer but with a couple of useful tricks.
Firstly observe that $\displaystyle \frac1X + \frac1Y + \frac1Z \ge 3\sqrt[3]{\frac1{XYZ}}$ = $\displaystyle 3 \big/ \sqrt[3]{XYZ} \ge \frac{9}{X+Y+Z}$, using AM/GM twice.
In your case this says that $\displaystyle \frac a{b+c} + \frac b{c+a} + \frac c{a+b} \ge 9\big/\big(\frac{b+c}a + \frac{c+a}b+\frac{a+b}c\big)$. I claim that this last sum is $\displaystyle \ge 6$. We can divide through by c and then we need to prove that $\displaystyle \frac{b+1}a + \frac{1+a}b+a+b \ge 6$. Fix b and treat as a function of a. It is minimised when the derivative = 0 and this is at $\displaystyle b^2 = a$ and the value is $\displaystyle a^2 + 1/a^2 + 2(a+1/a)$. A further application of AM/GM shows that $\displaystyle F+1/F \ge 2$ whatever F is, so the minimum is always $\displaystyle \ge 6$, as claimed.
I think that you have posted a wrong question and I have proved a wrong inequality.Originally Posted by Nichelle14
Let a=1, b=2 ,c=3
then
$\displaystyle \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{b+a}=\frac{1}{5} + \frac{2}{4} + \frac{3}{3} >3/2$
I have started a new thread 'Proof on inequality'in "pre algebra and algebra" asking for proof of the inequality$\displaystyle \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{b+a}\geq \frac{3}{2}$.
Could someone please help how I proved an inequality which does not exist?