Let \( a,b,c,d \in \mathbb{R^+}\). If \( \frac {1}{1+a} +\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d} = 3\). Prove that \(abcd \leq \frac{1}{81}\).

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TopNewestYou can do it like this too :

\[ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} + \frac{1}{1+d} = 3 \] \[ \frac{1}{1+a} = \frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d} \] Now using AM - GM \[ \frac{1}{1+a} = \frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d} \ge 3\sqrt[3]{\frac{bcd}{(1+b)(1+c)(1+d)}} \]

Thus ,

\[ \frac{1}{1+a} \ge 3\sqrt[3]{\frac{bcd}{(1+b)(1+c)(1+d)}} \] Similarly we get , \[ \frac{1}{1+b} \ge 3\sqrt[3]{\frac{acd}{(1+a)(1+c)(1+d)}} \] \[ \frac{1}{1+c} \ge 3\sqrt[3]{\frac{bcd}{(1+b)(1+a)(1+d)}} \] \[ \frac{1}{1+d} \ge 3\sqrt[3]{\frac{bcd}{(1+b)(1+c)(1+a)}} \]

Multiplying them all together , gives \[ 1 \ge 81 abcd \] As desired . :)

I have one more similar type of problem . You may like to try it .

Let \( x_1 , x_2 , \cdots , x_n \) be positive real numbers satisfying \[ \sum_{i=1}^{i=n} \frac{1}{x_i+1998} = \frac{1}{1998} \] Prove that , \[ \frac{(x_1x_2x_3.....x_n)^{\frac{1}{n}}}{n-1} \ge 1998 \] – Shivang Jindal · 4 years ago

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– Mridul Sachdeva · 4 years ago

Thanks..!Log in to reply

– Rizky Dermawan · 4 years ago

good solution... similar with my solution, but without trigonometric substitution (y)Log in to reply

try use trigonometric substitution, Let \( a = \ tan^{2} A, b = \ tan^{2} B, c = \ tan^{2} C, d = \ tan^{2} D\) and the equation becomes \( \cos^{2} A +\cos^{2} B+\cos^{2} C+\cos^{2} D=3 \) Observe that \( \cos^{2} A = \sin^{2} B + \sin^{2} C + \sin^{2} D\), then use AM-GM, we will get \( \cos^{2} A \geq 3 . (\sin B . \sin C . \sin D)^{\frac{2}{3}} \) Use the similar way to get \( \cos^{2} B, \cos^{2} C, \cos^{2} D \) Last, multiplying 4 inequalities that you have got, and you can find easily that \( abcd \leq \frac{1}{81}\) – Rizky Dermawan · 4 years ago

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This ain't nice but does the trick. Subject to that condition, there exist a maximum of \(abcd\) (We can do an argument that it is a function from a compact space to another compact space, so it must have a maximum.).

Suppose we have that maximum.Let \(x=\frac{1}{1+a}\) and similarly \(y,z\) and \(w\) for \(b,c,d\). Then \(a=\frac{1-x}{x}\) and \(\frac{1}{a}=\frac{x}{1-x}\). We have \(x+y+z+w=3\) and \(0\leq x,y,z,w\leq 1\) so \(x+y+2\geq x+y+z+w=3\) so \(2\geq x+y\geq 1\) so \(x+y=1+e\) with \(e\) in the interval \([0,1]\). (a variable, not the constant)

Now substitute \(y=1+e-x\) and \(\frac{1}{a}=\frac{x}{1-x}\) and similarly for \(b\) in \(\frac{1}{ab}\).Then

\[ \frac{1}{ab}=\frac{xy}{(1-x)(1-y)}=\frac{x(1+e-x)}{(1-x)(x-e)} =\frac{x-x^2+xe}{x-x^2+(x-1)e}=\frac{e}{x-x^2+(x-1)e}=1+\frac{e}{(1-x)(x-e)} \]

This function achieves maximum when \(1-x=x-e\), that is \(x=1+e-x=y\), that is \(a=b\). So, if we have \((a,b,c,d)\) that satisfy the hypothesis and is a maximum of \(abcd\) then it must have \(a=b\),else setting \(x'=y'=(x+y)/2\) would make another \((a',b',c,d)\) that satisfy the hypothesis with a strictly greater product. So, \(a=b\). In the same manner, \(b=c=d\). So, by hypothesis, \(a=\frac{1}{3}\). Thus the maximum of \(abcd\) is \((\frac{1}{3})^4=\frac{1}{81}\) – Diego Roque · 4 years ago

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One more solution , which is relied on again AM-GM inequality .

Substitute , \[ \frac{1}{1+a} = x , \frac{1}{1+b} = y , \frac{c}{1+c} = z , \frac{d}{1+d} = w \]. where \(x+y+z+w = 3 \). We get \[ a = \frac{1-x}{x} , b = \frac{1-y}{y} , c = \frac{1-z}{z} , d = \frac{1-w}{w} \] Now , We want to prove that \[ abcd \le \frac{1}{81} \] So it suffice to prove that , \[ 81(1-x)(1-y)(1-z)(1-w) \le xyzw \] since \[ 3 = x+y+z+w \] thus following inequality is equivalent to \[ (y+z+w-2x)(x+z+w-2y)(x+y+w-2z)(x+y+z-2w) \le xyzw \] Now we again do a obvious substitution , that is \( m = y+z+w-2x , n = x+z+w-2y , o = x+y+w - 2z , p = x+y+z-2w \) So , again the inequality is equivalent to , \[ mnop \le \frac{m+n+o}{3} \cdot \frac{n+o+p}{3} \cdot \frac{m+n+p}{3} \cdot \frac{m+o+p}{3} \] which is obvious by AM-GM :) – Shivang Jindal · 4 years ago

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Applying AM-HM inequality \(\large \frac{a+1 + b+1 + c+1 + d+1}{4} \ge \frac{4}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}.\) \( \Rightarrow a+b+c+d \ge \frac{4}{3}.\) Now applying AM-GM, \( \frac{a+b+c+d}{4} \ge (abcd)^{\frac{1}{4}}\) \( \Rightarrow \frac{1}{81} \ge abcd.\)

I think I am wrong in the last 2 steps.Please correct me... – Kishan K · 4 years ago

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– Shourya Pandey · 4 years ago

No. If \(a+b+c+d\) is greater than \(\frac { 4}{3}\) and \(4 (abcd)^{\frac {1}{4}}\), you cannot decide anything about the relation between the latter two.Log in to reply

– Cody Johnson · 4 years ago

That's what I was wondering!Log in to reply

HINT: One easy (but not so clever) way that works is to use Jensen's Inequality, and then use AM-GM. Although there are probably better ways to do this. – Jimmy Kariznov · 4 years ago

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– Mridul Sachdeva · 4 years ago

Thanks solved it.. But can you please tell me what are weights??..I studied wikipedia article on jensens inequality and it states something about weight associated with each number..I used the formula assuming equal weights and got the answer.. And we had to use the fact that \( y=\frac {1}{x+1}\) is a convex function?? Kindly can u please explain it to me in detail..Log in to reply

\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=3⇒\frac{1}{1+a}=(1-\frac{1}{1+b})+(1-\frac{1}{1+c})+(1-\frac{1}{1+d})=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}.Applying AM-HM inequality:\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\frac{1}{1+a}≥3\sqrt[3]{\frac{bcd}{(b+1)(c+1)(d+1)}}⇒\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}≥\frac{81abcd}{(a+1)(b+1)(c+1)(d+1}⇒ abcd≤\frac{1}{81} – Truong Nguyen Ngoc · 4 years ago

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Applying AM-HM inequality:

\[\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\frac{1}{1+a}≥3\sqrt[3]{\frac{bcd}{(b+1)(c+1)(d+1)}}⇒\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}≥\frac{81abcd}{(a+1)(b+1)(c+1)(d+1)}⇒ abcd≤\frac{1}{81}\] – Cody Johnson · 4 years ago

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either try Jensen's inequality or try to take a>b>c>d, and try to bound the value of a. – Utkarsh Mehra · 4 years ago

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\( 3 = \frac{3}{4} + \frac{3}{4} + \frac{3}{4} + \frac{3}{4} \)

\( \frac{3}{4} = \frac{1}{1 + a} = \frac{1}{1 + b} = \frac{1}{1 + c} = \frac{1}{1 + d} \)

\( \frac{1}{\frac{4}{3}} = \frac{1}{1 + a} = \frac{1}{1 + b} = \frac{1}{1 + c} = \frac{1}{1 + d} \)

\( \frac{4}{3} = 1 + a = 1 + b = 1 + c = 1 + d \)

So, \(a = b = c = d = \frac{1}{3} \)

Then, \( abcd = (\frac{1}{3}) ^ {4} \)

\( abcd = \frac{1}{81} \) – Alvin Willio · 4 years ago

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– Eric Edwards · 4 years ago

I think this comment gets at the intuition that the symmetric case of \(a=b=c=d\) will be critical. This is probably how the question was written in the first place, but it does not actually answer the question or provide a proof. Some sort of convexity inequality is clearly called for, and playing around with Jensen's and/or power means will eventually lead to the lovely arguments above.Log in to reply

proposed.

I agree, and in fact, Alvin's comment can be turned into a proof that one can easily do in one's own head. So all the downvotes were perhaps a bit unnecessary. The argument I'm going to sketch is actually a variant of what DiegoThe question can be rephrased as follows: maximize the function \(F(x,y,z,w) = f(x)+f(y)+f(z)+f(w)\), where \(f(t)=\log\left(\dfrac{1}{t}-1\right)\), subject to the constraints \(x+y+z+w=3\) and \(0<x,y,z,w<1\). One does need a small argument to show that \(F\)

doesachieve a maximum somewhere on this set, but once we know this, it follows by standard multi-variable calculus that at the point where the maximum is achieved, the gradient of \(F\) must be proportional to the vector \((1,1,1,1)\) (which is the gradient of the function \(H(x,y,z,w)=x+y+z+w-3\) that defines our hyperplane). This implies that \(x=y=z=w\) at the point where the maximum is achieved, and then Alvin's computation finishes the proof. – John Smith Staff · 4 years agoLog in to reply

The problem at first seems like it hinges on convexity, but the Lagrange multiplier method doesn't seem to involve it. Is convexity just a nice Olympiad-style shortcut or is it still operating in this solution, too? – Eric Edwards · 4 years ago

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same equation, and you are right that to show this one doesn't even need to compute any derivatives. Unfortunately, this doesn't immediately guarantee that they have to be thesame solutionof this equation.However, if you look at both the first and the

secondderivatives, you will see that in fact, \(x, y, z, w\) have to be the same because the equation \(f'(t)=k\) can have at most two solutions, and \(f''(t)\) is nonpositive for only one of those two solutions. (The point is that the second derivative can't be positive if you are at a point where a local max is achieved.)As far as convexity goes, it is certainly not just an Olympiad-style shortcut, but the problem is that I don't see a way to

immediatelyapply convexity here (and this is why all the other solutions that were posted required some additional calculations). For instance, if the function \(f(t)=\log(1/t-1)\) happened to be concave, we would be done immediately, but it just isn't: its second derivative has one sign on \((0,1/2)\) and the opposite sign on \((1/2,1)\). – John Smith Staff · 4 years agoLog in to reply