Titu’s Lemma

Titu’s Lemma states that for all positive real numbers

$x_1$, $x_2$, … , $x_n$ and $y_1$, $y_2$, … , $y_n$ it holds that

\[\frac{ x_1^2 } { y_1 } + \frac{ x_2 ^2 } { y_2 } + \cdots + \frac{ x_n ^2 } { y_n } \geq \frac{ (x_1 + x_2 + \cdots+ x_n ) ^2 } { y_1 + y_2 + \cdots + y_n }.\]

The equality holds if there exists $k$, such that $y_i=k x_i$ for all $i=1,2,…,n$.

 

Problem (Ireland, 2000)

Let $a$, $b$, $c$, $d$ be positive real numbers such that $a+b+c+d=1$. Show that

$$ \frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \geq \frac{1}{2} $$

Solution

Let us put

$$\left(x_1,x_2,x_3,x_4 \right)=\left( a,b,c,d \right)$$

$$\left(y_1,y_2,y_3,y_4 \right)=\left( a+b,b+c,c+d,d+a \right)$$

 
Let us now apply the Titu’s Lemma:

$$ \frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \geq $$

$$ \frac{ (a+b+c+d)^2}{2(a+b+c+d)} = \frac{1}{2} $$

which is what needed to be proved.

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