Proving Inequality for Positive Real Numbers

[anemone]

For positive real numbers $a,\,b,\,c$, prove the inequality:

\(\displaystyle a + b + c ≥ \frac{a(b + 1)}{a + 1} + \frac{b(c + 1)}{b + 1}+ \frac{c(a + 1)}{c + 1}\)

 

[anemone]

Hint:

Spoiler

Use substitution skill and let $x=a+1,\,y=b+1,\,z=c+1$.

 

[anemone]

My solution:

Spoiler

Let $x,\,y$ and $z$ be positive real such that $x=a+1$, $y=b+1$ and $z=c+1$, we see that by applying the AM-GM inequality to the sum of $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$ yields $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\ge 3$, so $-3\ge -\dfrac{x}{y}-\dfrac{y}{z}-\dfrac{z}{x}$. Next we add $(x+y+z)$ to both sides of the inequality, that gives $(x-1)+(y-1)+(z-1)≥ \left(x-\dfrac{x}{z}\right)+\left(y-\dfrac{y}{x}\right)+\left(z-\dfrac{z}{y}\right)=\dfrac{x(z-1)}{z}+\dfrac{y(x-1)}{x}+\dfrac{z(y-1)}{y}$, upon rearranging and back substituting we see that we've proved \(\displaystyle a + b + c ≥ \frac{a(b + 1)}{a + 1} + \frac{b(c + 1)}{b + 1}+ \frac{c(a + 1)}{c + 1}\).