Titu’s Lemma

Original Post.

Titu’s Lemma
It is just a variant of Cauchy Schwartz inequality. It states that for a, b > 0 and x, y\in\mathbb{R} we will have:
\frac {x^2}{a} + \frac {y^2}{b}\ge\frac {(x + y)^2}{a + b}
Generalisation
We have for x_i\in \mathbb{R} and a_i > 0 (i = 1, 2, \cdots n):
\sum_{i = 1}^{n} \frac {x_i^2}{a_i}\ge \frac {(\sum_{i = 1}^{n}x_i)^2}{\sum_{i = 1}^{n} a_i}
Hint for proof
Just use Cauchy

Applications
Using Titu’s lemma one can solve a variety of problems.A list of such problems:

1.

Prove that for positive reals a, b, c we have:
\frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}\ge \frac {3}{2} (NESSBITT)

2.

For positive unequal reals a, b, c we have:
\frac {2}{x + y} + \frac {2}{y + z} + \frac {2}{z + x} > \frac {9}{x + y + z}

3.

Prove that
\frac {a^2 + b^2}{a + b} + \frac {b^2 + c^2}{b + c} + \frac {c^2 + a^2}{c + a}\ge a + b + c for all reals a, b, c

4.

a, b, c are positive numbers such that abc = 1 Prove that:
\frac {1}{c^3(a + b)} + \frac {1}{a^3(b + c)} + \frac {1}{b^3(c + a)}\ge\frac {3}{2}

5.

For positive reals a, b, x, y, z we have:
\frac {x}{ay + bz} + \frac {y}{az + bx} + \frac {z}{ax + by}\ge\frac {3}{a + b}

6.

a, b, c, d, e > 0 Prove that
\frac {a}{c + b}+\frac {b}{c + d}+\frac {c}{d + e} + \frac {d}{e + a} + \frac {e}{a + b}\ge \frac {5}{2}

7.

x, y, z > 0 Prove that:
\frac {x^2}{(x + y)(z + x)} + \frac {y^2}{(y + z)(x + y} + \frac {z^2}{(z + x)(y + z)}\ge \frac {3}{4}

8.

Let ab + bc + ca = \frac {1}{3} for a, b, c > 0. Prove that:
\frac {a}{a^2 - bc + 1} + \frac {b}{b^2 - ca + 1} + \frac {c}{c^2 - ab + 1}\ge\frac {1}{a + b + c}

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