$\large P=\frac{1}{2+x+yz}+\frac{1}{2+y+xz}+\frac{1}{2+z+xy}$
If $x,y$ and $z$ are positive reals satisfying $4(x+y+z)=3xyz$, find the maximum value of $P$.

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@Laurent Michael
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They can but I am pretty much sure I have learned just enough in my 10th grade. I specialize in my field later. Thanks for the information!

Now the important thing to note here is that in P the variables are in the denominator. Also it is understood that if the denominator is big,the faction is small and vice versa. For the maximum value of P the denominators should be the smallest.

Now if we apply AM-GM inequality to 2+ x+yz
We get,

2+x+yz ≥(6)³√2 Six times cube root of 2

This will be the same for the other terms

So P =1/(6³√2 ) + 1/(6³√2 ) + 1/(6³√2 )

= 1/(2³√2) 1 divided by two times cube root of two

Easy Math Editor

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## Comments

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TopNewestI think I've got it! The answer is 3/8.

$\text{Let's switch variables to a, b and c, where, }$ $a = \frac{1}{yz}, b = \frac{1}{xz}, c = \frac{1}{xy} \\$ $\text{And P becomes, after substituting,} \\ P = \dfrac{a}{2a + \frac{a^2}{\sqrt{abc}} + 1} + \dfrac{b}{2b + \frac{b^2}{\sqrt{abc}} + 1} + \dfrac{c}{2c + \frac{c^2}{\sqrt{abc}} + 1} \\ \text{The given condition becomes, } a + b + c = \frac{3}{4} \\~\\ \text{So much nicer, right? Now, AM-GM gives} \\ (abc)^{1/3} \leq \frac{(a + b + c)}{3} = \frac{1}{4} \\ \Rightarrow abc \leq \frac{1}{64} \Rightarrow \sqrt{abc} \leq \frac{1}{8} \Rightarrow 8a^2 \leq \frac{a^2}{\sqrt{abc}} \\~\\ \text{As this term and similar ones are in the denominator,} \\ P \leq \dfrac{a}{8a^2 + 2a + 1} + \dfrac{b}{8b^2 + 2b + 1} + \dfrac{c}{8c^2 + 2c + 1} \\~\\ \\~\\ \text{Now, let's look at one of these terms, } \\ \dfrac{a}{8a^2 + 2a + 1} = \dfrac{2a}{16a^2 + 4a + 2} = \dfrac{2a}{(4a - 1)^2 + 12a + 1} \leq \dfrac{2a}{12a + 1} = \dfrac{1}{6} \cdot \left(1 - \dfrac{1}{12a + 1}\right) \\~\\ \\~\\ \text{But, by Cauchy-Schwarz (Titu's Lemma), } \\ \dfrac{1^2}{12a + 1} + \dfrac{1^2}{12b + 1} + \dfrac{1^2}{12c + 1} \geq \dfrac{(1 + 1 + 1)^2}{12(a + b + c) + 3} = \dfrac{9}{9 + 3} = \dfrac{3}{4} \\~\\ \text{Thus, } \\ P \leq \dfrac{1}{6} \cdot \left(3 - \left(\dfrac{1}{12a + 1} + \dfrac{1}{12b + 1} + \dfrac{1}{12c + 1}\right) \right) \leq \dfrac{1}{6} \cdot (3 - \dfrac{3}{4}) = \dfrac{3}{8} \Rightarrow P \leq \dfrac{3}{8} \\ \text{And, equality holds everywhere, if } a = b = c = \frac{1}{4} \Rightarrow x = y = z = 2$

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I feel that it's correct.

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What is the level of this question? What is the minimum grade we should have studied in to understand this question?

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Grade 10 mathematical olympiad

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Oh...Thanks!!

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Here is my solution. It has more of basic maths.

By applying AM-GM Inequality

xyz≥6

x+y+z≥8

Now the important thing to note here is that in P the variables are in the denominator. Also it is understood that if the denominator is big,the faction is small and vice versa. For the maximum value of P the denominators should be the smallest.

Now if we apply AM-GM inequality to 2+ x+yz We get,

2+x+yz ≥(6)³√2 Six times cube root of 2

This will be the same for the other terms

So P =1/(6³√2 ) + 1/(6³√2 ) + 1/(6³√2 )

= 1/(2³√2) 1 divided by two times cube root of two

≈0.396

And 0.396 > 0.375 which is 3/8

There you go bro...:) :)

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As pointed out by @Gurīdo Cuong, you "proved" incorrectly that $P \geq \approx 0.396$.

In particular, since $P ( 2, 2, ,2 ) = 3.75 < 0.396$, your solution is wrong.

Note: I do not know how you proved that $x + y + z \geq 8$, which is false like in the case of $x = y = z = 2$.

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I've thought of this solution before, but you see the equality holds when $x=y=z=xy=yz=xz=2$, which can't happened

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3/8

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$x=y=z=2$

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How do you know that?

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