How can we prove the inequality challenge for positive real numbers?

[anemone]

Let $a,\,b$ and $c$ be positive real numbers, prove that

\(\displaystyle \frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le \frac{3}{4}\).

 

[Euge]

Here is my solution.

Spoiler

Let $S = a + b + c$, $x = a/S$, $b = y/S$, and $z = c/S$. Then $x + y + z = 1$ and the above expression equals

$$\frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z}$$

which is the same as

$$\left(1 - \frac{1}{1 + x}\right) + \left(1 - \frac{1}{1 + y}\right) + \left(1 - \frac{1}{1 + z}\right)$$

or

$$3 - \left(\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z}\right)\tag{*}$$

By the arithmetic-harmonic mean inequality,

$$\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} \ge \frac{3^2}{(1 + x) + (1 + y) + (1 + z)} = \frac{9}{3 + (x + y + z)} = \frac{9}{3 + 1} = \frac{9}{4}$$

with equality if and only if $x = y = z$. Hence, the expression $(*)$ is no greater than $3 - 9/4 = 3/4$, with equality if and only if $x = y = z$ (which is equivalent to the condition $a = b = c$).

 

[Evobeus]

Let $a,\,b$ and $c$ be positive real numbers, prove that

\(\displaystyle \frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le \frac{3}{4}\).

Spoiler

Simple expansion does the tricky part and you are precisely done

 

[anemone]

Here is my solution.

Spoiler

Let $S = a + b + c$, $x = a/S$, $b = y/S$, and $z = c/S$. Then $x + y + z = 1$ and the above expression equals

$$\frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z}$$

which is the same as

$$\left(1 - \frac{1}{1 + x}\right) + \left(1 - \frac{1}{1 + y}\right) + \left(1 - \frac{1}{1 + z}\right)$$

or

$$3 - \left(\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z}\right)\tag{*}$$

By the arithmetic-harmonic mean inequality,

$$\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} \ge \frac{3^2}{(1 + x) + (1 + y) + (1 + z)} = \frac{9}{3 + (x + y + z)} = \frac{9}{3 + 1} = \frac{9}{4}$$

with equality if and only if $x = y = z$. Hence, the expression $(*)$ is no greater than $3 - 9/4 = 3/4$, with equality if and only if $x = y = z$ (which is equivalent to the condition $a = b = c$).

Very well done, Euge!(Cool) And thanks for participating!

My solution:

Spoiler

Let $x=a+b,\,y=b+c$ and $z=a+c$, this gives \(\displaystyle a=\frac{x-y+z}{2},\, b=\frac{x+y-z}{2}\), and \(\displaystyle c=\frac{y+z-x}{2}\) and the LHS of the intended inequality becomes

\(\displaystyle \begin{align*}\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}&=\frac{x-y+z}{2(x+z)}+\frac{x+y-z}{2(x+y)}+\frac{y+z-x}{2(y+z)}\\&=\frac{(x+z)-y}{2(x+z)}+\frac{(x+y)-z}{2(x+y)}+\frac{(y+z)-x}{2(y+z)}\\&=\frac{3}{2}-\frac{1}{2}\left(\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z}\right)\\&\le \frac{3}{2}-\frac{1}{2}\left(\frac{3}{2}\right),\,\,\text{since, by Nesbitt's inequality}\,\,\frac{y}{x+z}+\frac{z}{x+y}+\frac{x}{y+z}\ge \frac{3}{2}\\&=\frac{3}{4}\,\,\text{(Q.E.D.)} \end{align*}\)

Spoiler

Simple expansion does the tricky part and you are precisely done

Hi Evobeus!

Spoiler

I was wondering how the expansion would help to tackle the problem because for me, I always avoid the expansion (unless it is the last resort) in solving the inequality problems, could you post your entire solution here so I could learn something from it?