cn2_71828182846 wrote: Can someone explain a solution to USAMO 2001 problem 3? I am fine on the left inequality, but I'm a little hazy on the right-hand one. Thanks for assistance. Let a, b, c, be :ge: 0 and satisfy a :^2: + b :^2: + c :^2: + abc = 4. Show that 0 :le: ab + bc + ca - abc :le: 2. I'm not so sure but I'll give it a shot. -abc = a :^2: + b :^2: + c :^2: - 4 0 :le: ab + bc + ca + a :^2: + b :^2: + c :^2: - 4 :le: 2 4 :le: ab + bc + ca + a :^2: + b :^2: + c :^2: :le: 6 4 :le: [(a+b):^2: + (b+c):^2: + (a+c):^2:]/2 :le: 6 8 :le: (a+b):^2: + (b+c):^2: + (a+c):^2: :le: 12 That's all I've gotten so far, I'll work on it later.

I think I got the first part If one of a,b,c is 0, we are done. So assume a,b,c >0 By AM-HM(or cauchy): \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c} By the equation, a, b, c :le: 2. So a+b+c can be at most 6. Therefore \frac{9}{a+b+c} \geq 1 and \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 1 multiply both side by abc gives us the first inequality.

For the left side, just notice that not all of the variables can be >1, so WLOG assume a<=1. Then, bc(1-a)+a(b+c) >= 0 since each term is non-negative. It's the right side that is difficult. I have a solution, but it isn't entirely mine so I feel guilty posting it.

All I have for the second inequality: a^2+b^2 \geq 2ab b^2+c^2 \geq 2bc a^2+c^2 \geq 2ac add all together and divide by 2 a^2+b^2+c^2 \geq ab+bc+ac so ab+bc+ac+abc \leq 4 ab+bc+ac-abc \leq 4-2abc I will try to do figure it out later.

following off beta's work (i'm not quite sure if this works): The last inequality holds true if 2abc \ge 2 abc \ge 1 \sqrt[3]{abc} \ge 1 By GM-HM, we can substitute to get \displaystyle \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \ge 1 3 \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c} 3abc \ge ab+bc+ca From cauchy or AM-GM, we find a^2+b^2+c^2 \ge ab+bc+ca So we substitute again 3abc \ge a^2+b^2+c^2 Sorta guessing, but I think it can be proved that as long as the condition is satisfied, this inequality holds true, but i dont know how to do that.

paladin8 wrote: 3abc \ge ab+bc+ca From cauchy or AM-GM, we find a^2+b^2+c^2 \ge ab+bc+ca So we substitute again 3abc \ge a^2+b^2+c^2 That' not true, you can't substitute like that.

Yes, it works. If this is proved: 3abc \ge a^2+b^2+c^2 and this is true: a^2+b^2+c^2 \ge ab+bc+ca then this is true: 3abc \ge ab+bc+ca which is what we wish to prove.

ok, i see. I thought you proved 3abc \ge ab+bc+ca with these two steps, not realizing that you are working backwards a^2+b^2+c^2 \ge ab+bc+ca 3abc \ge a^2+b^2+c^2

beta wrote: All I have for the second inequality: a^2+b^2 \geq 2ab b^2+c^2 \geq 2bc a^2+c^2 \geq 2ac add all together and divide by 2 a^2+b^2+c^2 \geq ab+bc+ac so ab+bc+ac+abc \leq 4 ab+bc+ac-abc \leq 4-2abc I will try to do figure it out later. I don't think my idea is going to lead anywhere. (It's too strong.) To prove this statement to be true, we need abc \ge 1. However, by applying AM-GM to a^2 + b^2 + c^2 + abc = 4, \displaystyle{{a^2 + b^2 + c^2 + abc}\over{4}} \ge \sqrt[4]{a^3b^3c^3} From which we obtain abc \le 1. Hmm, thinking on then... Paladin: your steps don't hold. Try a = 2, b = c = 0.

Some more work One of a, b, c must be < or =1, one must be > or =1. Let a=1+x, let b=1-y. , where x, y \leq 1 so (1+x)(1-y)c \leq 1 (1+x)^2+(1-y)^2+c^2+(1+x)(1-y)(c)=4 since x^2+y^2+c^2-xyc \geq 0 because xy+yc+xc-xyc=xy+yc+xc(1-y) \geq 0 so we have (2+c)(1+x-y) \leq 4

Quote: Paladin: your steps don't hold. Try a = 2, b = c = 0. I assumed a, b, c > 0 because it is trivial to prove if any are 0 as noted by beta earlier on.

Let a = 1.999, b = c = 0.01. It still doesn't work.

that doesnt satisfy the original condition, but i have also found a counterexample, (1.99, .1, .1)

It doesn't?

1.999 :^2: +.01 :^2: +.01 :^2: +1.999*.01*.01 = 3.9964009.

Close enough. In any case, beta's assertion is an inequality that can't be proven. So we have to think about another way.

To prove this problem. We have to prove that, $ab+bc+ca-abc \leq 2 =\frac{a^2+b^2+c^2+abc}{2}\\ \Longleftrightarrow 2(ab+bc+ca) \leq a^2+b^2+c^2+3abc$

Well, let's see... By AM-GM, we have $a^2+b^2+c^2+abc\ge 4\sqrt[4]{(abc)^3}\Rightarrow 4\ge 4\sqrt[4]{(abc)^3}\Rightarrow 0\le abc\le 1$ (since $a,b,c\ge 0$). From there, we have $4\ge a^2+b^2+c^2\ge 3$. By AM-GM, $ab+ac+bc\ge 3\sqrt[3]{(abc)^2}$, so $ab+ac+bc\le 3$. All that's left to prove is that $ab+ac+bc\le 3abc$, but I'm not sure how to do that.

By the symmetry of the given equation, without loss of generality, $0\leq c\leq b\leq a,$ we have $4=a^2+b^2+c^2+abc\leq a^2+a^2+a^2+a\cdot a\cdot a\cdot\Longleftrightarrow a^3+3a^2-4\geq 0$ $\Longleftrightarrow (a-1)(a+2)^2\geq 0\Longleftrightarrow a\geq 1$,similarly we have $b\geq 1, c\geq1.$ Thus $4-2(ab+bc+ca-abc)=(a^2+b^2+c^2+abc)-2(ab+bc+ca-abc)$ $=a^2+b^2+c^2-2(ab+bc+ca)+3abc$ $\geq 3abc-(ab+bc+ca)\ (\because a^2+b^2+c^2\geq ab+bc+ca)$ $=abc+abc+abc-ab-bc-ca=ab(c-1)+bc(a-1)+ca(b-1)\geq 0$ $\Longleftrightarrow ab+bc+ca-abc\leq 2.$ Q.E.D.

Another way to solve the problem : We may take $a=\frac{2x}{\sqrt{(x+y)(x+z)}},b=\frac{2y}{\sqrt{(y+z)(y+x)}},c=\frac{2z}{\sqrt{(z+x)(z+y)}}$ where $x,y,z$ are positive real numbers.

For the first inequality, observe that by AM-GM it is enough to show that $ab+bc+ca \ge 3 \cdot (abc)^{\frac{2}{3}} \ge abc$. That is $27 \ge abc$, which is clear. For the second inequality, write $a=2 \cos A, b=2 \cos B, c=2 \cos C$ where $A+B+C=\pi$, thanks to the hypothesis and the identity $\cos^2 A+\cos^2 B+\cos^2 C+2 \cos A \cos B \cos C=1$. Rewrite the desired as \[ \sum_{cyc} 2 \cos A \cos B + \sum_{cyc} \cos^2 A - 1 = \sum_{cyc} (\cos A + \cos B)^2 - 1 \le 2 \]Subtracting $\cos^2 A + \cos^2 B + \cos^2 C$ from both sides, this becomes \[ (\cos A + \cos B + \cos C)^2 \le \sin^2 A + \sin^2 B + \sin^2 C \]Now we will prove two trig identities. Lemma 1: $\sum_{cyc} \cos A \sin B \sin C = \frac{1}{2} (\sin^2 A + \sin^2 B + \sin^2 C)$ where $A+B+C = \pi$. Proof: \begin{align*} \cos A \sin B \sin C &= \frac{1}{2} \cos A (\cos (B-C) + \cos A) \\ &= \frac{1}{2} \cos^2 A + \frac{1}{4} (\cos(A+B-C)+\cos(A+C-B)) \\ &= \frac{1}{2} \cos^2 A - \frac{1}{4} (\cos 2B + \cos 2C) \\ &= \frac{1}{2} \cos^2 A + \frac{1}{2} \sin^2 B + \frac{1}{2} \sin^2 C - \frac{1}{2} \\ &= \frac{1}{2} (\sin^2 B + \sin^2 C - \sin^2 A) \end{align*}Sum cyclically. Lemma 2: $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ where $A+B+C = \pi$. Proof: \begin{align*} \sin 2A + \sin 2B + \sin 2C &= 2 \sin (A+B) \cos (A-B) + \sin 2C \\ &= 2 \sin C (\cos (A-B) + \cos C) \\ &= 2 \sin C (\cos(A-B) - \cos (A+B)) \\ &= 4 \sin A \sin B \sin C \end{align*}As a simple consequence, we have that \begin{align*} \sum_{cyc} \frac{\cos A}{\sin B \sin C} &= \frac{\sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C} \\ &= \frac{\sin 2A + \sin 2B + \sin 2C}{2 \sin A \sin B \sin C} \\ &= 2 \end{align*} Now we apply Cauchy to finish the proof, combined with Lemmas 1 and 2: \[ \sin^2 A + \sin^2 B + \sin^2 C = (\sum_{cyc} \cos A \sin B \sin C)(\sum_{cyc} \frac{\cos A}{\sin B \sin C}) \ge (\cos A + \cos B + \cos C)^2 \]We are done.

v_Enhance wrote: Case 3: Now suppose that $a=b \neq c$. That means $\lambda = \frac{1}{2-c}$ Since we have that \[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]The result quoted seems to point to the fact that we need $(\left[ 2\lambda - 1 \right] - \lambda c)$ to equal zero, which confuses me. Isn't $(a-b)(\left[ 2\lambda - 1 \right] - \lambda c)$ already zero because $a=b$? Also, how does this work: v_Enhance wrote: Now \[ 2a+2c = a+b+2c = \lambda (2c+ab) = \frac{1}{2-c} (2c+a^2) \]which implies \[ 4a+4c-2a^2-2ac = 2c+a^2 \] Should it be $\lambda=\frac{1}{2-a}$? A final more general question about process, what was the motivation for letting having the condition for $U$ be $a^2+b^2+c^2<1000$? Why not let it be say $999$ or $1001$?. Is there anything specific about $1000$, or is it just a matter of choosing a number that is sufficiently large?

$\lambda = \frac{1}{2-a}$ was meant in both places. Thanks, I'll fix it. There's nothing special about $1000$ of course: as you say any large number suffices. The point is to capture the constraint set in some compact set, for which the boundary ($a^2+b^2+c^2=1000$ in this case) is a non-issue.

If $a,b,c\ge0$ such that $a^2+b^2+c^2+abc=4$, then $$a+b+c\le\sqrt{8+abc}\qquad$$

v_Enhance wrote: OK let's practice bashing. The left-hand side of the inequality is trivial; just note that $\min \left\{ a,b,c \right\} \le 1$. Hence, we focus on the right side. We use Lagrange Multipliers. Define \[ U = \left\{ (a,b,c) \mid a,b,c > 0 \text{ and } a^2+b^2+c^2 < 1000 \right\}. \]This is an intersection of open sets, so it is open. Its closure is \[ \overline U = \left\{ (a,b,c) \mid a,b,c \ge 0 \text{ and } a^2+b^2+c^2 \le 1000 \right\}. \]Hence the constraint set \[ \overline S = \left\{ \mathbf x \in \overline U : g(\mathbf x) = 4 \right\} \]is compact, where $g(a,b,c) = a^2+b^2+c^2+abc$. Excellent. Define \[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \]It's equivalent to show that $f \le 6$ subject to $g$. Over $\overline S$, it must achieve a global maximum. Now we consider two cases. If $\mathbf x$ lies on the boundary, that means one of the components is zero (since $a^2+b^2+c^2=1000$ is clearly impossible). WLOG $c=0$, then we wish to show $a^2+b^2+ab \le 6$ for $a^2+b^2=4$, which is trivial. Now for the interior $U$, we may use the method of Lagrange Multipliers. Consider a local maximum $\mathbf x \in U$. Compute \[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \]and \[ \nabla g = \left<2a+bc, 2b+ca, 2c+ab\right>. \]Of course, $\nabla g \neq \mathbf 0$ everywhere, so introducing our multiplier yields \[ \left<2a+b+c,a+2b+c,a+b+2c\right> = \lambda \left<2a+bc,2b+ca,2c+ab\right>. \]Note that $\lambda \neq 0$ since $a,b,c > 0$. Subtracting $2a+b+c = \lambda(2a+bc)$ from $a+2b+c = \lambda(2b+ca)$ implies that \[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]We can derive similar equations for the others. Hence, we have three cases. Case 1: If $a=b=c$, then $a=b=c=1$, and this satisfies $f(1,1,1) \le 6$. Case 2: If $a$, $b$, $c$ are pairwise distinct, then we derive $a = b = c = 2 - \lambda^{-1}$, contradiction. Case 3: Now suppose that $a=b \neq c$. That means $\lambda = \frac{1}{2-a}$ (of course our conditions force $c < 2$). Now \[ 2a+2c = a+b+2c = \lambda (2c+ab) = \frac{1}{2-a} (2c+a^2) \]which implies \[ 4a+4c-2a^2-2ac = 2c+a^2 \]meaning (with the additional note that $a \neq 1$) we have \[ c = \frac{3a^2-4a}{2-2a}. \]Note that at this point, $c > 0$ forces $1 < a < \frac 43$. The constraint $a^2+b^2+c^2+abc=4 \iff c^2 + + a^2c + (2a^2-4) = 0$ now gives \[ \left( 3a^2-4a \right)^2 + a^2 \left( 3a^2-4a \right)\left( 2-2a \right) + \left( 2a^2-4 \right)\left( 2-2a \right)^2 = 0. \]Before expanding this, it is prudent to see if it has any rational roots. A quick inspection finds that $a=2$ is such a root (precisely, $16-32+16=0$). Now, we can expand and try to factor: \begin{align*} 0 &= -6a^5 + 31a^4 - 48a^3 + 8a^2 + 32a - 16 \\ &= (a-2)(-6a^4 + 19a^3 - 10a^2 - 12a + 8) \\ &= (a-2)^2 \left( -6a^3+7a^2+4a-4 \right) \\ \intertext{Seeing a cubic, we desperately try every rational root and miraculously discover} &= (a-2)^2 (2-3a)(2a^2-a-2). \end{align*}The only root $a$ in the interval $\left( 1,\tfrac43 \right)$ is $a = \frac 14 \left( 1 + \sqrt{17} \right)$. You can guess what comes next -- write \[ c = \frac{a(3a-4)}{2-2a} = \frac{1}{8} \left( 7 - \sqrt{17} \right) \]and \[ f(a,b,c) = 3a^2+2ac+c^2 = \frac{1}{32} \left( 121 + 17\sqrt{17} \right) . \]This is the last critical point, so we're done once we check this is less than $6$. This follows from the inequality $17^3 < (6 \cdot 32 - 121)^2$; in fact, we actually have \[ \frac{1}{32} \left( 121 + 17\sqrt{17} \right) \approx 5.97165. \]This completes the solution. In case 3 , can anyone tell how c<2 is concluded?

Anyone ??

$c\ge2 \Longrightarrow c^2 \ge 4 \Longrightarrow a^2 + b^2 + c^2 + abc >4,$ contradiction.

cn2_71828182846 wrote: Let $a, b, c \geq 0$ and satisfy \[ a^2+b^2+c^2 +abc = 4 . \]Show that \[ 0 \le ab + bc + ca - abc \leq 2. \] 2017 Preparation for juniors, Romania: Let $a, b, c > 0$ and satisfy \[ a^2+b^2+c^2 = 3. \]Show that \[ ab + bc + ca - abc \leq 2. \]

Let $a \leq 1$ and $b,c \geq 1$. Now, obviously $(b-1)(c-1) \geq 0$ and: $ab+bc+ca-abc=a(b+c)+bc(1-a) \geq 0$ We will solve quadratic equation $a^2+b^2+c^2+abc-4=0$ for $a$, so: $a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$. And we will use this in second equation: $ab+bc+ca-abc=-abc+ab+ac-a+a+bc=-a(b-1)(c-1)+a+bc \leq a+bc=\frac{bc+\sqrt{(4-b^2)(4-c^2)}}{2}$ By Cauchy-Schwarz inequality we see that: $\frac{bc+\sqrt{(4-b^2)(4-c^2)}}{2} \leq \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)}}{2}=\frac{\sqrt{16}}{2}=2$ So, $0 \leq ab+bc+ca-abc \leq 2$

It is well-known that we can substitute $a=2\cos A,b=2\cos B,c=2\cos C$ for a triangle $ABC$. Observe that \[ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}\ge abc\]because $abc\le 27$, which shows the first part of the inequality. For the latter part, rewrite it as showing \[3\ge \sum_{\mbox{cyc}}(\cos A+\cos B)^2.\]Note since triangle $ABC$ is not obtuse, we can let $a^2=y+z,b^2=z+x,c^2=x+y$ for some nonnegative $x,y,z$, so the problem amounts to showing \[3\ge \sum_{\mbox{cyc}}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2 = 2\sum_{\mbox{cyc}}\frac{x^2}{(x+y)(x+z)}+2\sum_{\mbox{cyc}}\frac{yz}{(y+z)\sqrt{(x+z)(x+y)}}.\]By rearranging, the problem amounts to checking \[6xyz+3\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}x^2(y+z)+2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]By cancelling, it is equivalent to show \[6xyz+\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]But by AM-GM, the latter expression is at most \[\sum_{\mbox{cyc}}(2xyz+y^2z+z^2y),\]so done.

For the lower bound, write $0 \le ab + bc + ca - abc$ as \[0\le -a((b-1)(c-1)-1)+bc.\]This is clearly true since $(b-1)(c-1)-1$ must be nonpositive. Now we deal with the upper bound. WLOG, we can let $a-1$ and $b-1$ have the same sign. Then $(a-1)(b-1)=ab-b-a+1 \ge 0 \implies ab+1 \ge a+b$. Multiplying by $c$ and subtracting $abc$, we get $c \ge ac+bc-abc$. Adding $ab, $ \[c+ab \ge ac+bc+ab-abc.\]This means to prove the upper bound it suffices to prove $c+ab \le 2$. Assume FTSOC that $c+ab>2$. Then, \begin{align*} 4 &=a^2+b^2+c(c+ab) \\ &\ge 2ab+c(c+ab) \\ &> 2ab+2c \end{align*} which implies that $2>ab+c,$ contradiction. Hence, $c+ab \le 2$ and we are done.

My solution: We know that $min\{a,b,c\}\leq1$ so wlog any two of $a,b,c$ can be $\geq1$ or $\leq1$ So let $a\leq 1$ and $b,c\geq 1$ so we have \[ab+bc+ca-abc=a(b+c)+bc(1-a)\geq 0\]So lower bound is proved.So now we need to prove upper bound so \[a^2+b^2+c^2+abc=4\]and we know that $b^2+c^2\geq 2bc$ by AM-GM so now put this up\[a^2+2bc\leq 4\implies bc(2+a)\leq 4-a^2=(2+a)(2-a)\\\implies bc\leq 2-a\]now lets see what we have until now, $\bullet$ $(1-b)(1-c)\geq 0$ for both the cases $\bullet$ $bc\leq 2-a$ So now begin the sol lets see what we have \[ab+bc+ca-abc\leq ab+2-a+ca-abc=2-a(1-b-c+bc)=2-a(1-b)(1-c)\leq 2\]So we are done now !

Part 1: $0\leq ab+bc+ca.$ Notice $$\min (a,b,c) \leq 1 \implies ab+bc+ca \geq bc \geq abc.$$Equality when $a=b=0, c=2.$ Q.E.D. Part 2: $ab+bc+ca-abc\leq 2.$ WLOG, $a\leq 1, c\geq 1,$ then $$abc+b \geq ab+bc \implies b(a-1)(c-1)\geq 0.$$We have to prove $2\geq ac+b.$ Notice: $$a^2+c^2+b(ac+b)=4 \implies 2ac+b(ac+b) \leq 4 \implies (b+2)(ac+b-2) \leq 0 \implies ac+b \leq 2.$$Equality when $a=b=c=1.$ Q.E.D.

Set $a=2\cos A,b=2\cos B,c=2\cos C$ for $\triangle ABC$ acute. Then the stronger lower bound $ab+bc+ca-3abc\ge 0$ follows from Erdos-Mordell from the circumcenter of $ABC$ to its tangential triangle. If $\triangle ABC$ has side lengths $x,y,z$ we can write $a,b,c$ in terms of $x,y,z$ by Law of Cosines. We can check that \[abc+2-ab-bc-ca=\sum_{\text{cyc}}bc\cdot\frac{(y-z)^2}{2yz}\]which is nonnegative so we are done.