MHB Inequality Challenge: Prove $\ge 0$ for All $a,b,c$

AI Thread Summary
The inequality $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ is proposed for proof with positive real numbers $a$, $b$, and $c$. The discussion emphasizes the need to demonstrate that each term in the sum is non-negative. Participants engage in exploring various approaches, including algebraic manipulations and potential applications of known inequalities. The goal is to establish the validity of the inequality under the given conditions. The thread highlights the collaborative effort to solve this mathematical challenge.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ holds for all positive real $a,\,b$ and $c$.
 
Mathematics news on Phys.org
anemone said:
Prove $$\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}\ge 0$$ holds for all positive real $a,\,b$ and $c$.
my solution:
by using $AP\geq GP$ we have:

$\dfrac{a-\sqrt {bc}}{a+2b+2c}+\dfrac{b-\sqrt {ca}}{b+2c+2a}+\dfrac{c-\sqrt {ab}}{c+2a+2b}>
\\

\dfrac{a-{(b+c)/2}}{2a+2b+2c}+\dfrac{b-{(c+a)/2}}{2a+2b+2c}+\dfrac{c-{(a+b)/2}}{2a+2b+2c}=
0\\$
equality holds when $a=b=c$
 
Last edited:
Albert said:
my solution:
by using $AP\geq GP$ we have:

$\dfrac{a-\sqrt {bc}}{a+2b+2c}+\dfrac{b-\sqrt {ca}}{b+2c+2a}+\dfrac{c-\sqrt {ab}}{c+2a+2b}>
\\

\dfrac{a-{(b+c)/2}}{2a+2b+2c}+\dfrac{b-{(c+a)/2}}{2a+2b+2c}+\dfrac{c-{(a+b)/2}}{2a+2b+2c}=
0\\$
equality holds when $a=b=c$

Thanks Albert for participating!

Here is my solution:

$$\begin{align*}\frac{a-\sqrt{bc}}{a+2b+2c}+\frac{b-\sqrt{ca}}{b+2c+2a}+\frac{c-\sqrt{ab}}{c+2a+2b}&\ge \frac{a-\frac{b}{2}-\frac{c}{2}}{a+2b+2c}+\frac{b-\frac{c}{2}-\frac{a}{2}}{b+2c+2a}+\frac{c-\frac{b}{2}-\frac{a}{2}}{c+2a+2b}\\&\ge \frac{a-\frac{b}{2}-\frac{c}{2}}{3\sqrt{a^2+b^2+c^2}}+\frac{b-\frac{c}{2}-\frac{a}{2}}{3\sqrt{a^2+b^2+c^2}}+\frac{c-\frac{b}{2}-\frac{a}{2}}{3\sqrt{a^2+b^2+c^2}}\\&= \frac{1}{3\sqrt{a^2+b^2+c^2}}\left(a-\frac{b}{2}-\frac{c}{2}+b-\frac{c}{2}-\frac{a}{2}+c-\frac{b}{2}-\frac{a}{2}\right)\\&=\frac{1}{3\sqrt{a^2+b^2+c^2}}\left(1+b+c-a-b-c\right)\\&=0\,\,\,\,\text{Q.E.D.}\end{align*}$$

The first step follows from the AM-GM inequality that says $$\frac{b+c}{2}\ge \sqrt{bc}$$.

The second step follows from the Cauchy-Schwarz inequality that tells $$a+2b+2c\le\sqrt{1+2^2+2^2}\sqrt{a^2+b^2+c^2}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top