Negative discriminant and b+c nonzero not sufficient....So?

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ https://artofproblemsolving.com/community/c6h1634552p10320803

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 1 Let $b+c=p$, $a+c=q$ and $a+b=r$. Then: $px^2+qx+r=0$ has not real roots means that $q^2<4pr$. WLOG, we can assume that $p>0$ and $r>0$ and we have: $$3a(a+b+c) - 4ac+b^2 = 2 (p^2 - p q - 3 p r + 2 q r + 4 r^2)=f(q)$$Then $f$ is a linear function of $q$, so it is enough to check for extreme values of $q$: For $q=q_{max}=2\sqrt{pr}$, we get: \[ f(q_{max}) = (\sqrt{p}+\sqrt{r})^2(\sqrt{p}-2\sqrt{r})^2 \geq 0\]For $q=q_{min}=-2\sqrt{pr}$, we get: \[ f(q_{min}) = (\sqrt{p}-\sqrt{r})^2(\sqrt{p}+2\sqrt{r})^2 \geq 0\]Conclusion $f(q)\geq 0$ meaning that: $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 2 WLOG, we can assume that: $a =1$, and let $t =c-1-2b$. The equation doesn't have any root is equivalent to: \[t^2 \leq 8(b^2+b)\]We need to prove that: \[ t \leq b^2+b+2\]So it is enough to prove that: \[ 8(b^2+b) \leq (b^2+b+2)^2\]which is true beccause: \[ (b^2+b+2)^2-8(b^2+b) = (b-1)^2(b+2)^2 \geq 0 \]Done. Solution 3 If $a=0$, the problem is obvious. So let's assume $a$ different from 0 and let $t=3a(a+b+c)-4ac+b^2$ and we need to prove $t\geq 0$. We have: \[t=(2a+b)^2-a(a+b+c)\]So if $a(a+b+c) <0$ then obviously $t\geq0$. So, let's assume $a(a+b+c) \geq 0$. Let $P(x)=a((b+c)x^2+(a+c)x+(a+b))$ and $Q(x)=a(a+b+c)(x+\frac{1}{2})^2-a^2(x+\frac{b}{2a})^2$ and we have: \[P(x)=Q(x)+t\]Because $Q(-\frac{1}{2})\leq 0$ and $Q(-\frac{b}{2a})\geq 0$ then there exist $x_{0}$ such that $Q(x_{0})=0$. On the other hand $P$ doesn't have any real roots and $P(1)=2a(a+b+c)>0$ meaning that $P(x)>0$ for all $x$ real numbers. In particular for $x=x_{0}$ we have: \[P(x_{0})=Q(x_{0})+t=t> 0\]Done. Solution 4: We will prove it by absurd. Let's assume that $3a(a+b+c) < 4ac-b^2$ and we will prove that the equation $(b+c)x^2+(a+c)x+(a+b)=0$ has real roots. WLOG, we can suppose that $a>0$ then we have: $c > 3a+3b+\frac{b^2}{a}$. We will prove that: $$\Delta = (a+c)^2-4(a+b)(a+c) \geq 0$$. Let: \[f(x)=(a+x)^2-4(a+b)(a+x)\]We have: \[ f'(x) = 2(x-(a+2b))\]However: $3a+3b+\frac{b^2}{a} \geq a+2b$, then we have: \[\Delta = f(c) \geq f(3a+3b+\frac{b^2}{a})=\frac{(-2 a^2 + a b + b^2)^2}{a^2} \geq 0\]Contradiction. Solution 5: The problem is straightforward from this equality: \[(-b^2 - 3 a^2 - 3 ab + ac) (b^2 + a^2 - ab + ac)=-(a-b)^2(2a+b)^2+a^2((a+c)^2-4(a+b)(b+c)) < 0\]Or: $b^2+3a^2+3ab \geq ab-b^2-a^2$ then we conclude that: \[ ac \leq b^2+3a^2+3ab\]Or: \[ 4ac -b^2\leq3a(a+b+c)\]Done!

Use that the discriminant<0. And do some manipulations

Tuhin_Bose wrote: Use that the discriminant<0. And do some manipulations Did that! But perhaps we'd like to see your solution!

Tuhin_Bose wrote: Use that the discriminant<0. And do some manipulations it is what I did...

A more elegant solution: The key step is to add $ax^2+bx+c$ to get $\left(a+b+c\right)\left(x^2+x+1\right)=ax^2+bx+c$ Let $\left(a+b+c\right)\left(x^2+x+1\right)=f\left(x\right)$ and $ax^2+bx+c=g\left(x\right)$ As $\forall_{x\in R} \ f\left(x\right)\neq g\left(x\right)$ and they are both quadratics, either $f\left(x\right)>g\left(x\right)$ or $f\left(x\right)<g\left(x\right)$ for all $x$ Assume WLOG that $a+b+c$ is positive. Now $f\left(1\right)=3a+3b+3c,\ g\left(1\right)=a+b+c$ so $f\left(x\right)>g\left(x\right)$ for all $x$ It follows that $\min\left\{f\left(x\right)\right\}>\min\left\{g\left(x\right)\right\}$ $\min\left\{f\left(x\right)\right\}=\frac{3}{4}\left(a+b+c\right)$ If $a$ is positive, then $\min\left\{g\left(x\right)\right\}=\frac{4ac-b^2}{4a}$ and so $\frac{3}{4}\left(a+b+c\right)>\frac{4ac-b^2}{4a}$ or $3a\left(a+b+c\right)>4ac-b^2$ If $a$ is negative, then as $\max\left\{g\left(x\right)\right\}\ge g\left(1\right)$ we get $\frac{4ac-b^2}{4a}\ge a+b+c > \frac{3}{4}\left(a+b+c\right)$ which becomes the same inequality after multiplying by $a$ (the inequality flips as $a$ is negative)

Also @RagvaloD, could you please explain the grade system in Russia? (this question is grade 11 for example). I could not find any information on it online.

Very nice, Kaskade

Yes,very nice.Congratulations!

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 2 WLOG, we can assume that: $a =1$, and let $t =c-1-2b$. The equation doesn't have any root is equivalent to: \[t^2 \leq 8(b^2+b)\]We need to prove that: \[ t \leq b^2+b+2\]So it is enough to prove that: \[ 8(b^2+b) \leq (b^2+b+2)^2\]which is true beccause: \[ (b^2+b+2)^2-8(b^2+b) = (b-1)^2(b+2)^2 \geq 0 \]Done.

Kaskade wrote: Also @RagvaloD, could you please explain the grade system in Russia? (this question is grade 11 for example). I could not find any information on it online. 11th grade is the oldest grade in school.

11th grade - 18 and 19 years old.

WolfusA wrote: 11th grade - 18 and 19 years old. Not, 11th grade is 17-18 years old. Then at the age of 18, they enter the university

There doesn't occur equality case, as I analyzed #4.

WolfusA wrote: There doesn't occur equality case, as I analyzed #4. you are right

Kaskade wrote: As $f\left(x\right)=g\left(x\right)$ and they are both quadratics, either $f\left(x\right)>g\left(x\right)$ or $f\left(x\right)<g\left(x\right)$ for all $x$ I think you meant $\forall_{x\in R} \ f\left(x\right)\neq g\left(x\right)$

WolfusA wrote: Kaskade wrote: As $f\left(x\right)=g\left(x\right)$ and they are both quadratics, either $f\left(x\right)>g\left(x\right)$ or $f\left(x\right)<g\left(x\right)$ for all $x$ I think you meant $\forall_{x\in R} \ f\left(x\right)\neq g\left(x\right)$ Thanks, it is corrected.

Kaskade wrote: If $a$ is negative, then as $\max\left\{f\left(x\right)\right\}\ge f\left(1\right)$ we get $\frac{4ac-b^2}{4a}\ge a+b+c\ge\frac{3}{4}\left(a+b+c\right)$ which becomes the same inequality after multiplying by $a$ (the inequality flips as $a$ is negative) And here I observed it should be $\max\left\{g\left(x\right)\right\}\ge g\left(1\right)$

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 3 If $a=0$, the problem is obvious. So let's assume $a$ different from 0 and let $t=3a(a+b+c)-4ac+b^2$ and we need to prove $t\geq 0$. We have: \[t=(2a+b)^2-a(a+b+c)\]So if $a(a+b+c) <0$ then obviously $t\geq0$. So, let's assume $a(a+b+c) \geq 0$. Let $P(x)=a((b+c)x^2+(a+c)x+(a+b))$ and $Q(x)=a(a+b+c)(x+\frac{1}{2})^2-a^2(x+\frac{b}{2a})^2$ and we have: \[P(x)=Q(x)+t\]Because $Q(-\frac{1}{2})\leq 0$ and $Q(-\frac{b}{2a})\geq 0$ then there exist $x_{0}$ such that $Q(x_{0})=0$. On the other hand $P$ doesn't have any real roots and $P(1)=2a(a+b+c)>0$ meaning that $P(x)>0$ for all $x$ real numbers. In particular for $x=x_{0}$ we have: \[P(x_{0})=Q(x_{0})+t=t> 0\]Done.

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 4: We will prove it by absurd. Let's assume that $3a(a+b+c) < 4ac-b^2$ and we will prove that the equation $(b+c)x^2+(a+c)x+(a+b)=0$ has real roots. WLOG, we can suppose that $a>0$ then we have: $c > 3a+3b+\frac{b^2}{a}$. We will prove that: $$\Delta = (a+c)^2-4(a+b)(a+c) \geq 0$$. Let: \[f(x)=(a+x)^2-4(a+b)(a+x)\]We have: \[ f'(x) = 2(x-(a+2b))\]However: $3a+3b+\frac{b^2}{a} \geq a+2b$, then we have: \[\Delta = f(c) \geq f(3a+3b+\frac{b^2}{a})=\frac{(-2 a^2 + a b + b^2)^2}{a^2} \geq 0\]Contradiction.

RagvaloD wrote: $$(b+c)x^2+(a+c)x+(a+b)=0$$has not real roots. Prove that $$4ac-b^2 \leq 3a(a+b+c)$$ Solution 5: The problem is straightforward from this equality: \[(-b^2 - 3 a^2 - 3 ab + ac) (b^2 + a^2 - ab + ac)=-(a-b)^2(2a+b)^2+a^2((a+c)^2-4(a+b)(b+c)) < 0\]Or: $b^2+3a^2+3ab \geq ab-b^2-a^2$ then we conclude that: \[ ac \leq b^2+3a^2+3ab\]Or: \[ 4ac -b^2\leq3a(a+b+c)\]Done!

I'm referring to post 10. Take a look at this problem https://artofproblemsolving.com/community/c7h413602p2325352. btilm305 wrote: Suppose that $a, b, c, A, B, C$ are real numbers, $a \not= 0$ and $A \not= 0$, such that \[|ax^2+ bx + c| \le |Ax^2+ Bx + C|\]for all real numbers $x$. Show that \[|b^2- 4ac| \le |B^2- 4AC|\]

RagvaloD wrote: WolfusA wrote: 11th grade - 18 and 19 years old. Not, 11th grade is 17-18 years old. Then at the age of 18, they enter the university No, 16 or 17. In Russia children enter school at 6 and graduate at 17 (because at 18 they are drafted in the army unless enter a university).