MHB Proving Inequality for Positive Real Numbers

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The inequality states that for positive real numbers a, b, and c, the sum a + b + c is greater than or equal to the sum of three specific fractions involving a, b, and c. The fractions are structured to incorporate the values of a, b, and c along with their respective increments. A solution approach involves manipulating the terms to show that the left side dominates the right side. The discussion emphasizes the importance of algebraic manipulation and potentially applying known inequalities. A successful proof would confirm the validity of the inequality for all positive real numbers.
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For positive real numbers $a,\,b,\,c$, prove the inequality:

$$a + b + c ≥ \frac{a(b + 1)}{a + 1} + \frac{b(c + 1)}{b + 1}+ \frac{c(a + 1)}{c + 1}$$
 
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Hint:

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

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 $$a + b + c ≥ \frac{a(b + 1)}{a + 1} + \frac{b(c + 1)}{b + 1}+ \frac{c(a + 1)}{c + 1}$$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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