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Let $a,b,$ and $c$ be nonnegative real numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$0 \leq ab + ac + bc - abc \leq 2.$$

I tried using rearrangement to get $a^2+b^2+c^2+abc = 4 \geq ab+bc+ac+abc$. Then I just need to show that $0\leq ab + ac + bc - abc$ and $4-2abc \leq 2$. I am not sure if this method will work, though.

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This question is from the 30th USAMO 2001. Here you may find a solution:

30th USAMO 2001 question A3

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  • 1
    $\begingroup$ As I understand math.stackexchange.com/help/how-to-answer, an answer should contain the most relevant information in itself and not just a link (which may go offline in the future). $\endgroup$ – Martin R Jan 16 '16 at 17:57
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    $\begingroup$ Out from my post one can learn the following things: 1. There is an interesting math Olympiad that takes place in the US every year. 2. There is some origin to the problem, and maybe one will take the initiative to look older origin. 3. There is a good solution there that quoted and explained by the AoSP site. So in total, instead of copying that similar ideas into hear, one should open the link above and read and that after he/she tried to solve it by their own, of course. $\endgroup$ – Salech Rubenstein Jan 16 '16 at 18:10
  • $\begingroup$ I have two different solutions to this question, but one of them is very long and full of details. $\endgroup$ – Israel Meireles Chrisostomo Jan 2 '17 at 1:47
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I have a diferent solution for this question using a trigonometric substitution.Let:

$\\ \\ \displaystyle u=a\sqrt{\frac{2a+bc}{(2b+ac)(2c+ab)}};v=b\sqrt{\frac{2b+ac}{(2a+bc)(2c+ab)}};w=c\sqrt{\frac{2c+ab}{(2a+bc)(2b+ac)}} \\ \\$

This implies that:

$\\ \displaystyle a=2\sqrt{\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-uw}-1\right)}$ $\\ \displaystyle b=2\sqrt{\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-vw}-1\right)};$ $\\ \displaystyle c=2\sqrt{\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-vw}-1\right)};\\ $

Observe that:

$\\ \displaystyle a^{2}+b^{2}+c^{2}+abc=4\Rightarrow \left(\frac{1}{1-vw}-1\right)\left(\frac{1}{1-uw}-1\right)+ \left(\frac{1}{1-vw}-1\right)\left(\frac{1}{1-uv}-1\right)+\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-uv}-1\right)+ 2\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-vw}-1\right)=1\Rightarrow uv+vw+uw=1 $

Using that $\displaystyle uv+uw+vw=1$, our inequality is equivalent to:

. $\\ \\ \displaystyle \frac{2uv\sqrt{(u+w)(v+w)}}{(u+w)(v+w)(v+u)}+ \frac{2uw\sqrt{(v+w)(u+v)}}{(u+w)(v+w)(v+u)}+\frac{2vw\sqrt{(u+v)(u+w)}}{(u+v)(v+w)(u+w)} - \frac{4uvw}{(u+v)(v+w)(u+w)}\leq1\\ \\$

Let $\displaystyle a'=v+w,b'=u+w, c'=u+v$, observe that $\displaystyle a',b',c'$ will be sides of an arbitrary triangle.Let $\displaystyle S=\frac{a'+b'+c'}{2}$, we get $ \displaystyle u=S-a',v=S-b', w=S-c' $, and our inequality is equivalent to:

. \begin{equation} \frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'}+\frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'}+\frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}\leq 1 \tag{1} \end{equation}

We have to prove the inequality (1).Let's prove it, for that, consider that the square of every real number is positive, so we will have:

$ \\ \\ \displaystyle 0 \leq(\sqrt{a'}-\sqrt{b'})^2 \Rightarrow 2\sqrt{a'b'} \leq a'+b' \Rightarrow \frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'} \leq \frac{(a'+b')(S-a')(S-b')}{a'b'c'} \\ \\ $ $\displaystyle 0 \leq(\sqrt{a'}-\sqrt{c'})^2 \Rightarrow 2\sqrt{a'c'} \leq a'+c' \Rightarrow \frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'} \leq \frac{(a'+c')(S-a')(S-c')}{a'b'c'} \\ \\ $ $\displaystyle 0 \leq(\sqrt{b'}-\sqrt{c'})^2 \Rightarrow 2\sqrt{b'c'} \leq b'+c' \Rightarrow \frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'} \leq \frac{(b'+c')(S-b')(S-c')}{a'b'c'} \\ \\ $

Adding all the inequalities above and subtracting $\displaystyle \frac{4(S-a')(S-b')(S-c')}{a'b'c'} $ we get:

\begin{equation*} \frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'}+\frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'}+\frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'}-\frac{4(S-a')(S-b')(S-c')}{a'b'c'} \leq \end{equation*} \begin{equation} \frac{(a'+b')(S-a')(S-b')}{a'b'c'}+\frac{(a'+c')(S-a')(S-c')}{a'b'c'}+\frac{(b'+c')(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'} \tag{2} \end{equation}

Observe that:

$\\ \\ \displaystyle a'+b'=2\left(S-c'+\frac{c'}{2}\right); \displaystyle a'+c'=2\left(S-b'+\frac{b'}{2}\right);\displaystyle b'+c'=2\left(S-a'+\frac{a'}{2}\right)\\ \\$

Take the RHS of (2):

$\\ \\ \displaystyle \frac{(a'+b')(S-a')(S-b')}{a'b'c'}+\frac{(a'+c')(S-a')(S-c')}{a'b'c'}+\frac{(b'+c')(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}=$

$\\ \\ \displaystyle \frac{2\left(S-c'+\frac{c'}{2}\right)(S-a')(S-b')}{a'b'c'}+\frac{2\left(S-b'+\frac{b'}{2}\right)(S-a')(S-c')}{a'b'c'}+\frac{2\left(S-a'+\frac{a'}{2}\right)(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}=$

$\\ \\ \displaystyle \frac{(S-a')(S-b')}{a'b'}+\frac{(S-a')(S-c')}{a'c'}+\frac{(S-b')(S-c')}{b'c'} +\frac{2(S-a')(S-b')(S-c')}{a'b'c'} =\sin^{2}\frac{\alpha}{2}+ \sin^{2}\frac{\beta}{2}+ \sin^{2}\frac{\gamma}{2}+2\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \sin\frac{\gamma}{2}=1$

By the law of cosines.Replacing this result in RHS of (2).Thus (1) yields.

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The idea behind the solution is to make a trigonometric substitution and make the condition, which makes the inequality true, fall back into a valid trigonometric identity for angles of a triangle. See that we can do this using the property sine, cosine, tangent or cotangent have to be bijetors at intervals previously defined. Let's rewrite the condition so as to make it obvious. See:

$ \\ \\ \displaystyle a^{2}+b^{2}+c^{2}+abc=4 \Longrightarrow \frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{c^{2}}{4}+\frac{abc}{4}=1 \Longrightarrow \frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{c^{2}}{4}+\frac{2abc}{8}=1 \Longrightarrow $ $$ \left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}+2\left(\frac{a}{2}\right)\left(\frac{b}{2}\right)\left(\frac{c}{2}\right)=1 $$

Let $\displaystyle a=2sen\frac{\alpha}{2}$,$\displaystyle b=2sen\frac{\beta}{2}$,$\displaystyle c=2sen\frac{\gamma}{2}$, We fall back on the trigonometric identity: $$ sen^{2}\frac{\alpha}{2}+ sen^{2}\frac{\beta}{2}+sen^{2}\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\beta}{2}sen\frac{\gamma}{2}=1 $$

This trigonometric identity is valid when $ \displaystyle \alpha, \beta, \gamma $ are angles of a triangle. On the other hand, by doing this substitution in inequality, our inequality is rewritten as follows:

$ \\ \\ \displaystyle 4sen\frac{\alpha}{2}sen\frac{\beta}{2}+ 4sen\frac{\beta}{2}sen\frac{\gamma}{2}+4sen\frac{\alpha}{2}sen\frac{\gamma}{2}-8sen\frac{\alpha}{2}sen\frac{\beta}{2}sen\frac{\gamma}{2} \leq 2 \\ \\ 2sen\frac{\alpha}{2}sen\frac{\beta}{2}+ 2sen\frac{\beta}{2}sen\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\gamma}{2}-4sen\frac{\alpha}{2}sen\frac{\beta}{2}sen\frac{\gamma}{2} \leq 1 \\ \\ 2sen\frac{\alpha}{2}sen\frac{\beta}{2}+ 2sen\frac{\beta}{2}sen\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\gamma}{2} \leq 1+4sen\frac{\alpha}{2}sen\frac{\beta}{2}sen\frac{\gamma}{2} \\ \\ 2sen\frac{\alpha}{2}sen\frac{\beta}{2}+ 2sen\frac{\beta}{2}sen\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\gamma}{2} \leq cos\alpha+cos\beta+cos\gamma \\ \\ 2sen\frac{\alpha}{2}sen\frac{\beta}{2}+2sen\frac{\beta}{2}sen\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\gamma}{2} \leq \\ cos^{2}\frac{\alpha}{2}-sen^{2}\frac{\alpha}{2}+ cos^{2}\frac{\beta}{2}-sen^{2}\frac{\beta}{2}+ cos^{2}\frac{\gamma}{2}-sen^{2}\frac{\gamma}{2} \\ \\ 2sen\frac{\alpha}{2}sen\frac{\beta}{2}+2sen\frac{\beta}{2}sen\frac{\gamma}{2}+2sen\frac{\alpha}{2}sen\frac{\gamma}{2}+sen^{2}\frac{\alpha}{2} +sen^{2}\frac{\beta}{2}+sen^{2}\frac{\gamma}{2}\leq \\ cos^{2}\frac{\alpha}{2}+ cos^{2}\frac{\beta}{2}+ cos^{2}\frac{\gamma}{2} \\ \\ \left(sen\frac{\alpha}{2} +sen\frac{\beta}{2}+sen\frac{\gamma}{2} \right)^{2}\leq cos^{2}\frac{\alpha}{2}+ cos^{2}\frac{\beta}{2}+ cos^{2}\frac{\gamma}{2} \\ \\ cos^{2}\frac{\alpha}{2}+ cos^{2}\frac{\beta}{2}+ cos^{2}\frac{\gamma}{2} \geq \left(sen\frac{\alpha}{2} +sen\frac{\beta}{2}+sen\frac{\gamma}{2} \right)^{2} \\ \\ $

Inequality is equivalent to the above inequality. Now let's prove the inequality above. First, let S be the semi-perimeter and a, bec sides of a triangle. (Do not confuse the variables a, b, c of the statement with these variables a, b, C, which represent the sides of any triangle), then we get: $ \displaystyle \\ \\$

\begin{equation} sen\frac{\alpha}{2}=\sqrt{\frac{(S-b)(S-c)}{bc}}; sen\frac{\beta}{2}=\sqrt{\frac{(S-a)(S-c)}{ac}}; sen\frac{\gamma}{2}=\sqrt{\frac{(S-a)(S-b)}{ab}} \end{equation}

\begin{equation} cos\frac{\alpha}{2}=\sqrt{\frac{S(S-a)}{bc}}; cos\frac{\beta}{2}=\sqrt{\frac{S(S-b)}{ac}}; cos\frac{\gamma}{2}=\sqrt{\frac{S(S-c)}{ab}} \end{equation}

Consider an arbitrary triangle ABC and a circle inscribed in this triangle, so the points of tangency of the circle delimit two equal lines of two. Note that this statement is geometrically grounded, since if two lines tangentiate a circle and if these lines intercept In an outer point the same, then the distance from the point of intersection of the lines to the points of tangency are equal. Therefore, the following substitution is allowed $ \displaystyle a = x + y, b = y + z, c = x + z $. This transformation is known as Ravi transformation and can be useful in several other applications. , we'll have to:

$ \\ \\ \displaystyle S=\frac{a+b+c}{2}=\frac{(x+y)+(y+z)+(x+z)}{2}=x+y+z \\ \\$

We get:

$ \\ \displaystyle S-a=x+y+z-(x+y)=z$

$ \\ \displaystyle S-b=x+y+z-(y+z)=x$

$ \\ \displaystyle S-c=x+y+z-(x+z)=y $

Substituting above, we will have: \newpage

\begin{equation*} sen\frac{\alpha}{2}=\sqrt{\frac{xy}{(y+z)(x+z)}}; sen\frac{\beta}{2}=\sqrt{\frac{yz}{(x+y)(x+z)}}; sen\frac{\gamma}{2}=\sqrt{\frac{xz}{(x+y)(y+z)}} \tag{1} \end{equation*}

\begin{equation*} cos\frac{\alpha}{2}=\sqrt{\frac{z(x+y+z)}{(y+z)(x+z)}}; cos\frac{\beta}{2}=\sqrt{\frac{x(x+y+z)}{(x+y)(x+z)}}; cos\frac{\gamma}{2}=\sqrt{\frac{y(x+y+z)}{(x+y)(y+z)}} \tag{2} \end{equation*}

Consider the Cauchy-Schwarz inequality in 3 variables:

$\displaystyle (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\geq (a_1b_1+a_2b_2+a_3b_3)^2$

Making the substitution $\displaystyle a_{1}=\sqrt{xy},a_{2}=\sqrt{xz},a_{3}=\sqrt{yz}$ and $\displaystyle b_{1}=\sqrt{\frac{1}{(x+z)(y+z)}},b_{2}=\sqrt{\frac{1}{(x+y)(y+z)}},b_{3}=\sqrt{\frac{1}{(x+z)(x+y)}}$

You can see that it is worth the inequality below:

$\\ \\ \displaystyle (xy+xz+yz)\left( \frac{1}{(x+z)(y+z)}+\frac{1}{(x+y)(y+z)}+\frac{1}{(x+z)(x+y)}\right) \geq \\ \displaystyle \left( \sqrt{\frac{xy}{(x+z)(y+z)}}+\sqrt{\frac{xz}{(x+y)(y+z)}}+\sqrt{\frac{yz}{(x+z)(x+y)}}\right)^{2} \tag{Inequality A}$

$\\ $ Observe that:

$$ 2(xy+yz+xz)(x+y+z)=2(xy+yz+xz)(x+y+z) \Leftrightarrow $$

$$ (2xy+2yz+2xz)(x+y+z)=(xy+yz+xz)(2x+2y+2z) \Leftrightarrow$$

$$ ((xz+yz)+(xy+xz)+(xy+yz))(x+y+z)=(xy+yz+xz)((x+y)+(x+z)+(y+z)) \Leftrightarrow$$

$$ (z(x+y)+x(y+z)+y(x+z))(x+y+z)=(xy+yz+xz)((x+y)+(x+z)+(y+z)) $$ Dividing both sides of this equality by $ \displaystyle (x+y)(x+z)(y+z)$, we get:

$\\ \displaystyle \left( \frac{z}{(x+z)(y+z)}+\frac{x}{(x+z)(x+y)}+\frac{y}{(x+y)(y+z)}\right)(x+y+z)= \\ \displaystyle (xy+xz+yz)\left( \frac{1}{(x+z)(y+z)}+\frac{1}{(x+z)(x+y)}+\frac{1}{(x+y)(y+z)}\right) \\ $

We concludes the equality below:

$ \\ \\ \displaystyle \frac{z(x+y+z)}{(x+z)(y+z)}+\frac{x(x+y+z)}{(x+z)(x+y)}+\frac{y(x+y+z)}{(x+y)(y+z)}=\\ \displaystyle (xy+xz+yz)\left( \frac{1}{(x+z)(y+z)}+\frac{1}{(x+z)(x+y)}+\frac{1}{(x+y)(y+z)}\right) \\ \\ $

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Substituting this equality on the left side of `` Inequality A '', we get:

$ \\ \\ \displaystyle \frac{z(x+y+z)}{(x+z)(y+z)}+\frac{x(x+y+z)}{(x+z)(x+y)}+\frac{y(x+y+z)}{(x+y)(y+z)} \geq \\ \displaystyle \left( \sqrt{\frac{xy}{(x+z)(y+z)}}+\sqrt{\frac{xz}{(x+y)(y+z)}}+\sqrt{\frac{yz}{(x+z)(x+y)}}\right)^{2} \\ \\$

By doing the substitution (1) and (2) in the inequality above, finally we arrive at the result: $$cos^{2}\frac{\alpha}{2}+ cos^{2}\frac{\beta}{2}+ cos^{2}\frac{\gamma}{2} \geq \left(sen\frac{\alpha}{2} +sen\frac{\beta}{2}+sen\frac{\gamma}{2} \right)^{2} $$

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