Need help on this (full solution)

P=12+x+yz+12+y+xz+12+z+xy\large P=\frac{1}{2+x+yz}+\frac{1}{2+y+xz}+\frac{1}{2+z+xy} If x,yx,y and zz are positive reals satisfying 4(x+y+z)=3xyz4(x+y+z)=3xyz, find the maximum value of PP .

#Algebra

Note by P C
5 years, 3 months ago

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Comments

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

Let’s switch variables to a, b and c, where,  \text{Let's switch variables to a, b and c, where, } a=1yz,b=1xz,c=1xya = \frac{1}{yz}, b = \frac{1}{xz}, c = \frac{1}{xy} \\ And P becomes, after substituting,P=a2a+a2abc+1+b2b+b2abc+1+c2c+c2abc+1The given condition becomes, a+b+c=34 So much nicer, right? Now, AM-GM gives(abc)1/3≤(a+b+c)3=14⇒abc≤164⇒abc≤18⇒8a2≤a2abc As this term and similar ones are in the denominator,P≤a8a2+2a+1+b8b2+2b+1+c8c2+2c+1  Now, let’s look at one of these terms, a8a2+2a+1=2a16a2+4a+2=2a(4a−1)2+12a+1≤2a12a+1=16⋅(1−112a+1)  But, by Cauchy-Schwarz (Titu’s Lemma), 1212a+1+1212b+1+1212c+1≥(1+1+1)212(a+b+c)+3=99+3=34 Thus, P≤16⋅(3−(112a+1+112b+1+112c+1))≤16⋅(3−34)=38⇒P≤38And, equality holds everywhere, if a=b=c=14⇒x=y=z=2 \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

Ameya Daigavane - 5 years, 3 months ago

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I feel that it's correct.

Swapnil Das - 5 years, 3 months ago

3/8

Sumon Jose - 5 years, 3 months ago

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

Akhash Raja Raam - 5 years, 3 months ago

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Grade 10 mathematical olympiad

Son Nguyen - 5 years, 3 months ago

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Oh...Thanks!!

Akhash Raja Raam - 5 years, 3 months ago

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@Akhash Raja Raam – Also In Viet Nam student grade 9 who study specialized mathematics can solved this...

Son Nguyen - 5 years, 3 months ago

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@Son Nguyen – 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!

Akhash Raja Raam - 5 years, 3 months ago

x=y=z=2x=y=z=2

Son Nguyen - 5 years, 3 months ago

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How do you know that?

Pi Han Goh - 5 years, 3 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...:) :)

Surya Sharma - 5 years, 3 months ago

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

P C - 5 years, 3 months ago

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

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

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

Calvin Lin Staff - 5 years, 3 months ago
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