If given equation has five real roots, then the third derivative, which is quadratic polynomial has two real roots (just apply Rolle's theorem 3 times ). From the other hand, the inequality implies that its discriminant is negative.

leshik wrote: If given equation has five real roots, then the third derivative, which is quadratic polynomial has two real roots (just apply Rolle's theorem 3 times ). From the other hand, the inequality implies that its discriminant is negative. Is there a non-calculus solution?

The actual problem statement of USAMO 1983 #2 is: Prove that the roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$ cannot all be real if $2a^2 < 5b$.

Thanks. I just conjured up the solution with AM-GM. Note to Binomial-Theorem- EDIT THE PROBLEM STATEMENT

EDIT: I made a mistake in my solution. Will fix later.

Wait does @AopsUser101's solution work? Because in the last two step, it says that $5(a^2-2b) \ge a^2$, but how does that lead to $2a^2 \le 5b$?. And didnt you want to prove that $2a^2 \ge 5b$? Sorry, Im a bit confused

mynameisbob12 wrote: Wait does @AopsUser101's solution work? Because in the last two step, it says that $5(a^2-2b) \ge a^2$, but how does that lead to $2a^2 \le 5b$?. And didnt you want to prove that $2a^2 \ge 5b$? Sorry, Im a bit confused Yes, there was a problem. I’ll fix is later, as time allows.

My solution is pretty much the same as @AIME15, just wanted to elaborate on the AM-GM step.

Since $(r_1)^2+...+(r_5)^2 =a^2-2b$, by Cauchy we know that $(a^2-2b)(5) \ge a^2$,so $5b \le 2a^2$, so we have proved the contapositive, and so we are done

Let $r_1, r_2, r_3, r_4, r_5$ be the roots. Then by Vieta's, we have \[a = - \sum_{1 \le i \le 5}r_i\] and \[b = \sum_{1 \le i < j \le 5} r_ir_j.\] Then the inequality \[2a^2 < 5b\] becomes \[2\cdot\left(\sum_{1 \le i \le 5}r_i\right)^2 < 5 \cdot \left(\sum_{1 \le i < j \le 5} r_ir_j\right)\] or \[2\cdot\left(\sum_{1 \le i \le 5}r_i^2\right) < \sum_{1 \le i < j \le 5} r_ir_j.\] Now, assume for the sake of contradiction that none of the $r$'s are unreal. Then the squares of the $r_i$ are nonnegative. This means, we can apply AM-GM to get that \[\frac{r_i^2 + r_j^2}{2} \ge r_ir_j\] for all $1 \le i < j \le 5$. Summing this over all $i, j$, we get \[2\cdot\left(\sum_{1 \le i \le 5}r_i^2\right) \ge \sum_{1 \le i < j \le 5} r_ir_j\] which is a contradiction, as desired. $\square$

Funny problem. Rewrite the quintic as $$a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0.$$We will prove the contrapositive: if all roots to the quintic are real, then $2(a_4)^2\ge 5(a_3)$. Let the roots be $a,b,c,d,e$ and define the sums $P_k=a^k+b^k+c^k+d^k+e^k$. By Newton, \begin{align*}0&=a_5P_1+a_4\\&= a_5P_2+a_4P_1+a_3\cdot 2,\end{align*}but since $a_5=1$, we have $P_1=-a_4$ and $P_2=-a_4P_1-2a_3 = (a_4)^2-2a_3$. Therefore, $(a_4)^2=P_2+2a_3$ and we wish to show $$2(P_2+2a_3)\ge 5(a_3)\Rightarrow 2P_2\ge a_3.$$In expanded form, this is $$2(a^2+b^2+c^2+d^2+e^2)\ge ab+ac+ad+ae+bc+bd+be+cd+ce+de$$which is immediate by Muirhead since $(2,0,0,0,0)\succ (1,1,0,0,0)$.

Let $r_1, r_2, r_3, r_4, r_5$ be the roots; assume FTSOC they are real. The condition implies $$2(r_1+r_2+r_3+r_4+r_5)^2 < 5\sum_{\mathrm{sym}} r_1r_2 \iff 2\sum_{i=1}^5 r_i^2 < \sum_{\mathrm{sym}} r_1r_2,$$which is obviously absurd.

Denote the roots by $r_1$, $r_2$, \dots, $r_5$. By Vieta's, we have \[r_1+r_2+\dots+r_5=-a\qquad\text{and} \qquad\sum_{i<j}r_ir_j=b.\]Note that \[(r_1+r_2+\dots+r_5)^2=a^2=r_1^2+r_2^2+\dots+r_5^2 +2\sum_{i<j}r_ir_j=(r_1^2+r_2^2+\dots+r_5^2)+2b.\]If $r_1^2+r_2^2+\dots+r_5^2<0$, then the roots cannot all be real, so if $a^2<2b$, then the roots cannot be all real. $2a^2<5b$ is weaker than $a^2<2b$, which implies the result.

Suppose that the roots were all real. Call the roots $r_1, r_2, r_3, r_4, r_5$. Note that by Vieta's, \[ a^2 = \left(\sum_{i=1}^5 r_i\right)\text{ and } b = \left( \sum_{1\le i< j \le 5} r_i r_j \right) \]Using the fact that $a^2 + b^2 \ge 2ab$ for any reals $a,b$ (follows from the fact that $(a-b)^2 \ge 0$), we find that \begin{align*} 4 \sum_{i=1}^5 r_i ^2 = \sum_{1\le i < j \le 5} (r_i^2 + r_j^2) \ge 2 \left( \sum_{1\le i< j \le 5} r_i r_j \right) \\ \implies 2\sum_{i=1}^5 r_i ^2 \ge \left( \sum_{1\le i< j \le 5} r_i r_j \right),\\ \end{align*}Now, we have \begin{align*} 2a^2 = 2\left(\sum_{i=1}^5 r_i\right) \\ = 2 \sum_{i=1}^5 r_i ^2 + 4 \left( \sum_{1\le i< j \le 5} r_i r_j \right) \\ \ge 5 \left( \sum_{1\le i< j \le 5} r_i r_j \right) = 5b,\\ \end{align*}a contradiction. Therefore the roots cannot be all real.

Let $x_1,x_2,\dots,x_5$ be the roots of $p(x)=x^5+ax^4+bx^3+cx^2+dx+e$. We note that $p$ has $$\sum_{i=1}^5 x_i=-a\quad\text{and}\quad\sum_{\text{sym}} x_ix_j=b$$$$\sum_{i=1}^5 x_i^2=a^2-2b$$So in fact $2a^2<5b$ is equivalent to $4\sum_{i=1}^5 x_i^2<-\left(\sum_{i=1}^5 x_i\right)^2$ which immeditately implies $p$ cannot have $5$ real roots.

FTSOC, assume that all roots are real. Let these roots be WLOG $r_1\leq r_2\leq \dots \leq r_5$. Note that we have that \[a=r_1+r_2+r_3+r_4+r_5,\]and \[b=\sum r_xr_y,\]for all pairs of positive integers $(x,y)$ such that $1\leq x\leq y\leq 5$. By Muirhead's [$(2,0,0,0,0)$ majorizes $(1,1,0,0,0)$], we have that \[2(r_1^2+r_2^2+\dots+r_5^2)\geq b,\]and adding $4b$ to both sides gives us that \[2(r_1^2+r_2^2+\dots+r_5^2)+2(2b)\geq 5b \iff{} 2a^2\geq 5b,\]a contradiction, finishing the problem.

Assume that all of the roots are real. Denote them by $r_1, r_2, r_3, r_4, r_5$. Then, \[a = -\sum_{1 \le i \le 5} r_i\]\[b = \sum_{1 \le i < j \le 5} r_i r_j.\] Now, it suffices to prove \[2 \left ( \sum_{1 \le i \le 5} r_i \right )^2 \ge 5 \sum_{1 \le i < j \le 5} r_i r_j\] which is equivalent to \[2 \sum_{1 \le i \le 5} r_i^2 \ge \sum_{1 \le i < j \le 5} r_i r_j.\] This results from summing $\frac{r_i^2+r_j^2}{2} \ge r_ir_j$ over $1 \le i < j \le 5$. $\square$

Assume for the sake of contradiction all roots are real, denote them by $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ Then $$2a^{2}=2(x_{1}+x_{2}+x_{3}+x_{4}+x_{5})^{2}$$$$5b=5\sum_{\text{cyc}} x_{1}x_{2}$$If $$2a^{2} < 5b $$$$2(x_{1}+x_{2}+x_{3}+x_{4}+x_{5})^{2}<5\sum_{\text{cyc}} x_{1}x_{2}$$$$ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} <\sum_{\text{cyc}} x_{1}x_{2} <5\sum_{\text{cyc}} |x_{1}x_{2}| $$But, by AM-GM $$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq 5\sum_{\text{cyc}} |x_{1}x_{2}| $$A contradiction $\square$

Assume that they are all real, say $r_i$. We easily find that the condition implies, \begin{align*} 2\left( \sum_i r_i \right)^2 < 5\sum_{i < j} r_ir_j \end{align*}This expands as, \begin{align*} 2 \sum_i r_i^2 &< \sum_{i < j} r_ir_j\\ \end{align*}However note $r_i^2 + r_j^2 > 2r_ir_j$. Then over all pairs of $(i, j)$ we find, \begin{align*} 4 \sum_i r_i^2 &\geq 2 \sum_{i < j} r_ir_j\\ 2 \sum_i r_i^2 &\geq \sum_{i < j} r_i r_j \end{align*}contradiction. $\square$

Let $p$, $q$, $r$, $s$, and $t$ be the roots of the given. We can see from vieta's that $2a^2 = 2(p+q+r+s+t)^2$. We can also see that $5b = 5\sum_{sym} pq$. We want to prove that $2(p+q+r+s+t)^2 \geq 5\sum_{sym} pq$. Crossing out terms we get $2\sum_{sym} p^2 \geq \sum_{sym} pq$ which follows from Muirhead. $\blacksquare$

We will show that if all the roots are real, then we must have $2a^2\ge 5b$. Let the roots be real numbers $p,q,r,s,t$. Then we want to show that $$2(p+q+r+s+t)^2\ge 5(pq+pr+ps+pt+qr+qs+qt+rs+rt+st)$$or $$2p^2+2q^2+2r^2+2s^2+2t^2\ge pq+pr+ps+pt+qr+qs+qt+rs+rt+st.$$By AM-GM, $\frac{p^2+q^2}2\ge pq$, and the desired result follows.

We will use contradiction; assume that the roots are real. Call them $x_1, x_2, \dots, x_5$. Note that $$a = -x_1-x_2-\dots-x^5$$and $$b = x_1x_2+x_1x_3+\dots+x_4x_5.$$Since $2a^2<5b$, we know that $$2(x_1+x_2+\dots+x_5)^2<5(x_1x_2+x_1x_3+\dots+x_4x_5)$$or $$2(x_1^2+x_2^2+\dots+x_5^2+2(x_1x_2+x_1x_3+\dots+x_4x_5))<5(x_1x_2+x_1x_3+\dots+x_4x_5).$$Simplifying, we have $$2(x_1^2+x_2^2+\dots+x_5^2) < x_1x_2+x_1x_3+\dots+x_4x_5.$$This follows by Muirhead.

Instructively.....