The polynomial $ax^2+bx+c$ satisfies: $a&gt;0, a+b+c \ge 0, a-b+c \ge 0, a-c \ge 0$ and $b^2-4ac \ge 0$. Prove that the roots are real and that they belong to the interval $-1 \le x \le 1$    I started proving it this way: Here it is stated $b^2 - 4ac \ge 0$, therefore, the roots are real.    Now the roots are: $ -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$    From the inequalities $a-b+c \ge 0$ and $a-c\ge0$, we get $2a\ge b$ Therefore, $-\frac{b}{2a}\ge-1$ and $b \ge a \Rightarrow b &gt; 0$ Now I can see the other part of the root can be also shown to be greater than or equal 1. For that, I'll have to prove $4a^2\ge b^2-4ac$ Squaring both sides of  $2a\ge b$, I get $4a^2\ge b^2$ And from inequality $a-c \ge 0$ we get $a \ge c \Rightarrow 4a^2 \ge 4ac$ I don't know how to proceed from here. Is this the correct approach to prove it? If not, then how should I prove the statement?
121323
If we have:  $a&gt;0,a+b+c \ge 0,a-b+c \ge 0,a-c \ge 0,b^2-4ac \ge 0$  Then Prove that there exists two roots in $[-1,1]$.  We have $f(1),f(-1) \ge 0$ so if we could prove there is a number $c$ that $f(c)&lt;0$ and $-1&lt;c&lt;1$ we are done but I can't do that.  Edit:$f(x)=ax^2+bx+c$  Edit 2 I made a typo and corrected it now.Now using answer of @George Law It only remains to check the case $\Delta =0$ which is likely to cause problem with $a-c \ge o$ because we didn't use it.
62349
$$Q(x) = x^k + a_1x^{k+1}+ ... + a_nx^{k+n}$$ where $k,n$ are positive integers is a polynomial with real coefficients. I have to show that $Q(x)/x^k$ is strictly positive for all real x satisfying $$0&lt;|x|&lt;1/(1+\sum_{i=0}^n |a_i|)$$  Now the polynomial formed after division(which is legal since $x$ is non-zero) is the general $n$ degree polynomial $1+a_1x+...+a_nx^n.$ This approaches $1$ as $x$ approaches $0$. But how can I obtain the right hand part of the inequality? Furthermore, I tried substituting $x=1$ to obtain the right hand denominator but the it doesn't give the sum of the absolute values.
29094
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$   Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.   This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.
10631
Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant.
2904
Let $$P(x) = x^4+a_3x^3+a_2x^2+a_1x+a_0$$   $$P(x_0) = 0$$   $$P'(x_0) \not= 0$$   with $x_0$ and each $a_i$ real. Prove that $P(x)$ has a at least two real roots.   I can't figure why this is true.
2457
Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$  Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.  My idea:  I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.  Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$ I have to show that $$y^2-2xz\leq 3$$ I tried to prove it with the sign of trinom, but it doesn't work.
43018
Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.  Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$    Here's my idea:  $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  And I'm stuck here.  I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$  In the first case using Cauchy-Schwarz Inequality I prove that:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$  Now I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$   $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$   I need I don't know how to continue.  In the second case I tried proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and  $a^2 + b^2 + c^2 \ge a+b+c$  Using Cauchy-Schwarz Inequality I proved:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$  $a^2 + b^2 + c^2 \ge a+b+c$  But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$  So please help me with this problem.  P.S   My initial idea, which is proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  maybe isn't the right way to prove this inequality.
6895
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that   $a^2 + b^2 + c^2 \ge a + b + c$.   I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
23436
If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that  $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $  ( don't apply Schur's inequality )? Thanks.
92362
Question is to check :  For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .  the way in which i have proceeded is :  let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$  i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$  But, $a^3+a+c=0$. So, $-a^3-a=c$.  so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$  i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$  Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 &lt;0$ for any real number $a$  Thus, quadratic has no real root and so is the cubic $x^3+x+c$  I would like to know if this justification is sufficient and if this can be generalized.  I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)  can this be generalized to any odd degree polynomial (at least for some special cases)
45145
Is it true that  $$P_n(x):=1+ \sum_{m=1}^n\dfrac{x^m}{m!}$$ has no real root for even $n$ and exactly one real root for odd $n$? I can only prove that the polynomial cannot have any multiple roots. Am stuck , Please Help.
24930
It seems to be true that every real polynomial $p_n$ of degree $n$ can be factored in the following (not unique) way $$ p_n = \sum_{i=0}^n a_ix^i = s \left\{\prod_{i=1}^n(x-b_i)\right\} +t \tag{$\ast$}$$ with $a_i, s, b_i, t $ all in $\mathbb{R}$.  For example:   $x-1=(x-1)+0$ $x^2-5x+7=(x-2)(x-3)+1=(x-5)(x-0)+7$ $x^2+1 = (x-0) \cdot (x-0) +1$   I am having a hard time coming up with a rigorous proof. I have tried using the fact that every polynomial $p_n$ can be written as $(x-a)g(x)+b$, where $g$ is a polynomial function and $a$ is an arbitrary real number, but with no success. Another way it could be done is by expanding the right-hand side of $(\ast)$ and showing that the linear system of equations with the coefficients of matching powers of $x$ has always a real solution. But I feel there must be a more simple proof out there, e.g. by induction, if possible without the Fundamental Theorem of Algebra. Would be glad if someone could enlighten me!
92917
I need to prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational. Obviously, it's easy to use the rational root theorem to prove that there are not rational solutions to this equation, but i want a different approach. I saw a similar question here, but it didn't provide my "solution".  My try  $x^3+x+1=0$  $x^3+x=-1$  $x(x^2+1)=-1$  Here we have two options (product of two numbers to obtain a negative):  $1.$ $x \gt 0$ $∧$ $x^2+1 \lt 0$ (not possible in $\Bbb R)$  $2.$ $x \lt 0$ $∧$  $x^2+1 \gt 0$   So, we use $2.$ to prove that $x \ne 0$, then:  $x(x^2+1)=-1$  $x^2+1=\frac{-1}{x}$  $x^2=\frac{-1}{x}-1$  $x^2=-(\frac{x+1}{x})$  $x=\pm \sqrt {-(\frac{x+1}{x})}$  Then, $\frac{x+1}{x} \ge 0$ because $x \lt 0$ (the equality occurs when $x=-1$, but this doesn't satisfy the original polynomial equation) but here i'm missing the cases $-1 \lt x \lt 0$ (i don't know how to use this to prove that the root is irrational)  This implies that $2$ of the roots are complex, but there are $3$ roots to a third degree polynomial equation.  And here i'm stuck, because i don't know how to prove that the last solution is irrational.  Any hints?  Is there anyway to prove that the last root is irrational?  Is my proof good so far?
94331
Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.
40105
Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0&lt;a_0&lt;a_1&lt;\cdots&lt;a_n.$$  I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).  Any ideas?
105975
Let: $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ where $a_0a_n &lt; 0$ I have to prove that the Polynomial $P(x)$ has at least one positive root how can I prove it? Any ideas?
11804
The problem statement is:  Part(a)  Assume P(z) is a non-constant polynomial with all of its roots in some half plane H.  Show that the derivative P'(z) must also have all of its roots in H.  Part (b)  Show that the roots of P' always lie within the closed convex hull of the set of roots of P.  The hint given is:   After obvious reductions of the problem, look at the behavior of $\frac{P'}{P}$ on the complementary half-plane.  My thoughts:   So I think that, whatever half-plane H that we have to work with, we can always just look at $P(e^{i\theta}z)$, which has its zeros in the UHP, for a suitable rotation by angle theta.    Now we can just show that $P'(e^{i\theta}z)e^{i\theta}$ also has its roots in the UHP -- and then undoing the rotations shows that both P and P' had its roots in the same half-plane H.  Using the hint regarding looking at the complementary half-plane, we then want to show that P' does not vanish in the LHP (we already know P doesn't vanish in the LHP, by assumption.)    Again following the hint we look at P'/P in the LHP and show that it does not vanish.  I am stuck at this point.  How can I proceed?  And regarding part(b) what is a "convex hull"?  I am guessing that the answer to part (b) is immediate, after figuring out part(a).  Any ideas are welcome.  Thanks,