$x_i^2\geq 4(x_i-1)$ and AM-GM

math90 wrote: $x_i^2\geq 4(x_i-1)$ and AM-GM Thanks.

but how??

Let $x_1+x_2+x_3+......+x_{2011}=k$ and let the L.H.S of the inequality be $A$ (I am too lazy to type the whole thing ).Now apply titu's lemma on all the 2011 fractions to get: $A\geq \frac {k^2}{k-2011} $ Now if we prove $$\frac {k^2}{k-2011} \geq 8044$$we are done and$$\frac {k^2}{k-2011} \geq 8044$$is trivial to prove.

math90 wrote: $x_i^2\geq 4(x_i-1)$ and AM-GM But how ? ..I mean please explain the first statement about x^2 ...according to your statement.. It would yield x > 2 which is not.given (X>1 is given)..please explain

Note: $x^2\geq 4(x-1)\iff (x-2)^2\geq 0$

$$\displaystyle{\frac{x^2_1}{x_2-1}- 4x_1 + 4(x_2-1) +\ldots+\frac{x^2_{2011}}{x_1-1} -4x_{2011} +4 (x_1-1)} $$is sum of square number then $ x_1 = 2(x_2-1) , ... , x_{2011} = 2(x_1-1)$. Then $x_1=x_2= ... =x_{2011}=2$.

Let $s=x_1+x_2+\ldots +x_{2011}>2011,$ then by Cauchy-Schwarz Inequality, we have \[\frac{x^2_1}{x_2-1}+\frac{x^2_2}{x_3-1}+\frac{x^2_3}{x_4-1}+\ldots+\frac{x^2_{2010}}{x_{2011}-1}+\frac{x^2_{2011}}{x_1-1}\geqslant \frac{s^2}{s-2011} \geqslant 8044 \iff (s-4022)^2\geqslant 0\]which is true. Equality holds when $x_1=x_2=\ldots =x_{2011}=2. \quad \blacksquare$

sqing wrote: Let $\displaystyle {x_i> 1, \forall i \in \left \{1, 2, 3, \ldots, 2011 \right \}}$. Show that:$$\displaystyle{\frac{x^2_1}{x_2-1}+\frac{x^2_2}{x_3-1}+\frac{x^2_3}{x_4-1}+\ldots+\frac{x^2_{2010}}{x_{2011}-1}+\frac{x^2_{2011}}{x_1-1}\geq 8044}$$When the equality holds? Holder ineq for overkill . Here we assume that $x_{2011+k}=x_k$ for $k \le 2011$ Use holder ineq with $p=q=2$ and the secuences $X_i=\frac{x_i}{\sqrt{x_{i+1}-1}}$ , $Y_i=\sqrt{x_{i+1}-1}$ and $M_i=1$ : $$\sum_{i=1}^{2011} \frac{x_i^2}{x_{i+1}-1} \ge \frac{(\sum_{i=1}^{2011} x_i)^2}{\sum_{i=1}^{2011} x_{i+1}-2011}=\frac{(\sum_{i=1}^{2011} x_i)^2}{\sum_{i=1}^{2011} x_{i}-2011} \ge 8044$$And the last one holds using that for any real number $a>2011$ it follows that: $$\frac{a^2}{a-2011} \ge 8044 \implies (a-4022)^2 \ge 0 \; \text{and this is truth}$$Thus we are done