a_507_bc wrote: Let $a, b$ be reals with $a \neq 0$ and let $$P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b.$$Show that all roots of $P(x)$ are real and positive if and only if $a=b$. We will rather consider $P(x)=x^4-4x^3+(5+c)x^2-4cx+c$. If $c=1$, then $P(x)=(x-1)^4$ which has only positive real roots. Suppose $P$ has only positive real roots $r_1,…,r_4$ and let’s prove that $c=1$. Using Newton’s identities, we have $(p_1,p_2,p_3,p_4)=(4,2(3-c),4,2(c^2+8c-7))$ where $p_k=\sum r_i^k$. From CS, $4p_2\geq p_1^2 \implies c \leq 1$ CS again $4p_4\geq p_2^2 \implies (c-1)(c+23)\geq 0\implies c\geq 1$. So $c=1$

a_507_bc wrote: Let $a, b$ be reals with $a \neq 0$ and let $$P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b.$$Show that all roots of $P(x)$ are real and positive if and only if $a=b$. Viete’s formulae yield $\sum x_i=4=\sum \frac{1}{x_i}$, thus by CS and since $(\sum x_i) \cdot (\sum \frac{1}{x_i})=16$, $x_i=1$. The same formulae yield: $a=b$.

For some reason, I didn't think of the Cauchy-Schwarz solution during the exam, so I went with AM-GM. The forward direction is very straight forward. If $a=b$, then $P(x)=a(x-1)^4$ which has real and positive roots $x_1=x_2=x_3=x_4=1$ For the other direction, since the roots are all real and positive, we can apply AM-GM on them. By Vieta's formulae, $\sum{x_i} = 4 , \sum{x_ix_jx_k} = \frac{4b}{a} , \prod{x_i} = \frac{b}{a} $ So, by AM-GM we have that $\frac{\sum{x_i}}{4} \geq \sqrt[4]{\prod{x_i}} \\ \Longrightarrow 1 \geq \sqrt[4]{\frac{b}{a}} \\ \Longrightarrow \frac{b}{a} \leq 1$ and $\frac{\sum{x_ix_jx_k}}{4} \geq \sqrt[4]{\left(\prod{x_i}\right)^3} \\ \Longrightarrow \frac{b}{a} \geq \sqrt[4]{\left(\frac{b}{a}\right)^3} \\ \Longrightarrow \frac{b}{a} \geq \left(\frac{b}{a}\right)^\frac{3}{4} \\ \Longrightarrow \frac{b}{a} \geq 1$

Thus, combining both inequalities, we get that $\frac{b}{a} = 1$ or $a = b. &nbsp_\square$

Rukevwe wrote: For some reason, I didn't think of the Cauchy-Schwarz solution during the exam, so I went with AM-GM. I wouldn't consider this to be something negative. As I am very fond of telling the South African students during training: every single inequality that has ever been in the PAMO can be solved either with no named inequalities, or with AM-GM. (And not in some unnatural or convoluted way. i.e. I'm not claiming that it can be solved with AM-GM because it can be solved with Cauchy and Cauchy can be proven by AM-GM. I'm claiming that there is a natural solution that uses "only" AM-GM.) Of course this probably won't stay true forever.

We split the problem into two parts: $\bullet$ Assume that $P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b$ has $4$ roots $x_1,x_2,x_3,x_4\in \mathbb{R}^+$ By Vieta's formulas, we have$$\left\{ \begin{array}{l} {x_1} + {x_2} + {x_3} + {x_4} = 4\\ {x_1}{x_2}{x_3} + {x_1}{x_2}{x_4} + {x_1}{x_3}{x_4} + {x_2}{x_3}{x_4} = \dfrac{{4b}}{a}\\ {x_1}{x_2}{x_3}{x_4} = \dfrac{b}{a} \end{array} \right.$$$\Rightarrow x_1x_2x_3x_4.\left( {\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}} \right)=\dfrac{4b}{a}\Rightarrow \dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}=4,$ by Titu's lemma, we have$$4=\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}+\dfrac{1}{x_4}\geq \dfrac{16}{x_1+x_2+x_3+x_4}=4$$So the equality occurs if and only if $x_1=x_2=x_3=x_4=1\Rightarrow \dfrac{b}{a}=1\Rightarrow a=b.$ $\bullet$ Assume that $a=b,$ we have $P(x)=ax^4-4ax^3+6ax^2-4ax+a=a.(x-1)^4$ Then all roots of $P(x)$ are all equal to $1\in \mathbb{R}^+.$ Conclusion: all roots of $P(x)$ are real and positive if and only if $a=b$.

a_507_bc wrote: Let $a, b$ be reals with $a \neq 0$ and let $$P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b.$$Show that all roots of $P(x)$ are real and positive if and only if $a=b$. Let $k_1,k_2,k_3,k_4$ be the solutions of the polynomial $P(x)$ $$a(x-k_1)(x-k_2)(x-k_3)(x-k_4)=(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b.$$$$ax^4-a(\sum_{cyc} k)x^3+a(\sum_{cyc} k_1k_2)x^2-a(\sum_{cyc} k_1k_2k_3)x +a k_1k_2k_3k_4=ax^4-4ax^3+(5a+b)x^2-4bx+b$$$$\left\{ \begin{array}{l} {k_1} + {k_2} + {k_3} + {k_4} = 4\\ {k_1}{k_2}{k_3} + {k_1}{k_2}{k_4} + {k_1}{k_3}{k_4} + {k_2}{k_3}{k_4} = \dfrac{{4b}}{a}\\ {k_1}{k_2}{k_3}{k_4} = \dfrac{b}{a} \end{array} \right.$$$\implies k_1k_2k_3k_4.\left( {\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}+\dfrac{1}{k_4}} \right)=\dfrac{4b}{a}\implies \dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}+\dfrac{1}{k_4}=4$ or $b=0$ by Titu's lemma \[\frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }.\]$k_1k_2k_3k_4.\left( {\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}+\dfrac{1}{k_4}} \right)=\dfrac{4b}{a}\implies \dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}+\dfrac{1}{k_4}=4\ge\frac{(1+1+1+1)^2}{k_1+k_2+k_3+k_4}=4$ the equality holds if and only if $k_1=k_2=k_3=k_4=1$ so$$P(x)=a(x-1)^4=a(x^4-4x^3+6x^2-4x+1)$$$$=ax^4-4ax^3+6ax^2-4ax+1=ax^4-4ax^3+(5a+b)x^2-4bx+b\implies 6a=5a+b\implies a=b $$now testing for $b=0$ $$\boxed{\textbf{Claim}:P(x)=ax^4-4ax^3+5ax^2\text{has no real solutions}} $$it's easy to show that the claim is true let $P(x)=0$ $$ax^2(x^2-4x+5)=0$$so $a=0$ (refused) or $x^2-4x+5=0 \implies x=2+i \text{or} x=2-i$ hence we are done $\blacksquare$

The problem is equivalent to showing that the roots of the polynomial $x^4-4x^3+(5+k)x^2-4kx+k$ are all positive iff $k=1$. Indeed suppose the roots are $a,b,c,d>0$. By Vieta's formulas and AM-GM: $$1=\frac{a+b+c+d}4\geq (abcd)^{1/4}=k^{1/4}\implies k\leq 1$$and $$k=\frac{abc+abd+acd+bcd}4\geq (abcd)^{3/4}=k^{3/4}\implies k\geq 1$$so $k=1$. Conversely if $k=1$ then the polynomial is simply $(x-1)^4$.