biquadratic equation........

I think i have traced the solution ;)

okay..so at the end of the day i will present the solution. the solution is really awesome but very tricky as well.

Let the roots of the polynomial equation f(x) = x⁴+ ax³+ bx²+ax+c , be p,q,r,s

We have (1-b+c)² = f(i)f(-i) = (p²+1)(q²+1)(r²+1)(s²+1), where 'i 'is iota.

By CS inequality (p²+1)(q²+1)(r²+1)(s²+1) ≥ (1+ √|pqrs|)⁴ (*) = (1+ √|c|)⁴

Note that by vieta's theorem pqrs = c

Thus the inequality follows..

Edits: how to get the inequality (*)

By direct CS inequality

(p²+1)(q²+1)≥(1+|pq|)² and (r²+1)(s²+1)≥(1+|rs|)²

multiplying (p²+1)(q²+1)(r²+1)(s²+1)≥(1+|pq|)²(1+|rs|)²

Again by CS (1+|pq|)(1+|rs|)≥(1+ √|pqrs|)²

Ultimately we get (p²+1)(q²+1)(r²+1)(s²+1) ≥ (1+ √|pqrs|)⁴

@man111: Did u get the qstn at some place related to jee?

Great Shubhodip very Nice and best ever solution.

from old test papers of mtg

CS inequality: For reals a₁, a₂, b₁,b₂ the following inequality holds

(a₁²+b_1²)(a₂²+ b₂²)≥(a₁a₂+ b₁b₂)²

proof: simplify and rearrange and use the fact that No Square is Negative. also try to write the inequality in vector form.

$Let $\vec{A}=a_{1}\hat{i}+b_{1}\hat{j}$ and $\vec{B}=a_{2}\hat{i}+b_{2}\hat{j}$\\\\ Then $\vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos\;\phi\Leftrightarrow (\vec{A}.\vec{B})^2=|\vec{A}|^2|\vec{B}|^2cos^2\phi\leq |\vec{A}|^2|\vec{B}|^2$\\\\ $\left(a_{1}a_{2}+b_{1}b_{2}\right)^2\leq \left(a_{1}^2+b_{1}^2\right).\left(a_{2}^2+b_{2}^2\right)\Leftrightarrow \boxed{\boxed{\left(a_{1}^2+b_{1}^2\right).\left(a_{2}^2+b_{2}^2\right)\geq \left(a_{1}a_{2}+b_{1}b_{2}\right)^2}}$