<s>[INST] $a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $ If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)^2 \ge \frac{289}{4} $$
I tried expanding the entire LHS and I got 
$a^2 + b^2 + c^2 + d^2 + \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + 8$ on the LHS. Then I tried applying AM-GM to the first 4 terms and the last 4 terms and I got 
$4 \sqrt{abcd} + \frac{4}{\sqrt{abcd}}$ but that didnt seem useful.
Any hints?
Thank you in advance.
 [/INST] If you consider $a,b,c,d$ are all positive
According to cauchy-schwarz inequality
$(1+1+1+1)*((a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2)>=$
$(a+b+c+d+1/a+1/b+1/c+1/d)^2=$
$(a+b+c+d+4+b/a+a/b+c/a+a/c+d/a+a/d+c/d+d/c+c/b+b/c+d/b+b/d)^2>=(5+12)^2$[because $a/b+b/a>=2 and a+b+c+d=1$]
So we get.... $(a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2>= 289/4$.
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