6

If $p,q,r,s>0$ and $(p+q)(q+r)(r+s)(s+p)=16$ and $pq+qr+rs+sp=2$.

Then minimum value of $(p^2+q^2+r^2+s^2)$

Try: using Cauchy Schwarz Inequality

$(p^2+q^2+r^2+s^2)(q^2+r^2+s^2+p^2)\geq (pq+qr+rs+sp)^2=4$

But answer given as $6$. Could some help me to solve it. Thanks

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1 Answers1

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Write $p^2 + q^2 + r^2 + s^2 = X$. Then, $$2X + 2(pq+qr+rs+sp) = (p+q)^2 + (q+r)^2 + (r+s)^2 + (s+p)^2 = 2X + 4$$.

Write $a = p+q, b = q+r, c= r+s, d = s+p$, then by AM-GM on $a^2,b^2,c^2,d^2$ we see that $a^2+b^2+c^2+d^2 \geq 4 \sqrt[4]{(abcd)^2} = 16$.

Therefore, $2X+4 \geq 16$, and $X \geq 6$.

Equality is attained, precisely when all terms in the AM-GM are equal i.e. $p=r,q=s$, in which case $p+q = 2$ and $pq = \frac 12$. Solve to get $p,q = 1 \pm \frac 1{\sqrt 2}$, at which indeed $X= 6$ is obtained. Hence, the minimum value is $X = 6$.

Sarvesh Ravichandran Iyer
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  • Nice work! I wanted to use the fact that $a^2+b^2+c^2+d^2 ≥ 16$, but I didn't know the trick in the second line! – Toby Mak Mar 29 '18 at 13:23
  • I did not believe it when I saw it, because I thought if a question has been upvoted five times, then it must require some fair amount of effort to solve, but this worked a treat. Once that part was done : I must admit, I would have been in some doubt had the OP not mentioned that the answer was $6$, although it would have by equality in the AM-GM inequality. – Sarvesh Ravichandran Iyer Mar 29 '18 at 13:25