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Hari Shankar (9114)

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From Cauchy-Schwarz Theorem (in Titu's Lemma form) we have

$\frac{1}{m} + \frac{1}{m} + \frac{1}{n} + \frac{1}{k} \ge \frac{16}{2m+n+k}$

Thus we form three inequalities as below:

$\frac{2}{m} + \frac{1}{n} + \frac{1}{k} \ge \frac{16}{2m+n+k} \\ \\\frac{1}{m} + \frac{2}{n} + \frac{1}{k} \ge \frac{16}{m+2n+k} \\ \\\frac{1}{m} + \frac{1}{n} + \frac{2}{k} \ge \frac{16}{m+n+2k}$

Adding the above we get $4 \left(\frac{1}{m} + \frac{1}{n} + \frac{1}{k} \right) \ge 16 \left(\frac{1 }{2m+n+k} + \frac{1}{m+2n+k}+ \frac{1}{m+n+2k} \right)$

Now we are given that $\left(\frac{1}{m} + \frac{1}{n} + \frac{1}{k} \right) = 4$

Hence we obtain the inequality $\frac{1}{2m+n+k} + \frac{1}{m+2n+k} + \frac{1}{m+n+2k} \le 1$

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