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# Need help on this (full solution)

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

Note by Gurīdo Cuong
1 year, 2 months ago

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

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

I feel that it's correct. · 1 year, 1 month ago

What is the level of this question? What is the minimum grade we should have studied in to understand this question? · 1 year, 2 months ago

Oh...Thanks!! · 1 year, 2 months ago

Also In Viet Nam student grade 9 who study specialized mathematics can solved this... · 1 year, 2 months ago

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! · 1 year, 2 months ago

3/8 · 1 year, 2 months ago

$$x=y=z=2$$ · 1 year, 2 months ago

How do you know that? · 1 year, 2 months ago

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...:) :) · 1 year, 2 months ago

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 · 1 year, 2 months ago

Comment deleted Feb 23, 2016

I'm talking about his solution. The way he used AM-GM led to $$x=y=z=xy=yz=xz=2$$ when the equality holds · 1 year, 2 months ago

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 $$x = y = z = 2$$. Staff · 1 year, 2 months ago