21I think i have traced the solution ;)
$\textbf{The equation $\mathbf{x^4+ax^3+bx^2+ax+c=0,a,b,c\in\mathbb{R}$} }$\\\\ $\textbf{has all real roots . Then Prove that $\mathbf{\sqrt{|1-b+c|}\geq 1+\sqrt{|c|}}$}
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21okay..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?
1708Great Shubhodip very Nice and best ever solution.
from old test papers of mtg
21CS inequality: For reals a1, a2, b1,b2 the following inequality holds
(a12+b12)(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.
1708$Let $\vec{A}=a_{1}\hat{i}+b_{1}\hat{j}$ and $\vec{B}=a_{2}\hat{i}+b_{2}\hat{j}$\\\\ Then $\vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos\;\phi\Leftrightarrow (\vec{A}.\vec{B})^2=|\vec{A}|^2|\vec{B}|^2cos^2\phi\leq |\vec{A}|^2|\vec{B}|^2$\\\\ $\left(a_{1}a_{2}+b_{1}b_{2}\right)^2\leq \left(a_{1}^2+b_{1}^2\right).\left(a_{2}^2+b_{2}^2\right)\Leftrightarrow \boxed{\boxed{\left(a_{1}^2+b_{1}^2\right).\left(a_{2}^2+b_{2}^2\right)\geq \left(a_{1}a_{2}+b_{1}b_{2}\right)^2}}$
