# Prove the Inequality

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 $$abcd \leq \frac{1}{81}$$.

Note by Mridul Sachdeva
4 years, 11 months ago

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## Comments

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You 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 , $\frac{(x_1x_2x_3.....x_n)^{\frac{1}{n}}}{n-1} \ge 1998$

- 4 years, 11 months ago

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Thanks..!

- 4 years, 11 months ago

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good solution... similar with my solution, but without trigonometric substitution (y)

- 4 years, 11 months ago

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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 $$abcd \leq \frac{1}{81}$$

- 4 years, 11 months 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 $$(\frac{1}{3})^4=\frac{1}{81}$$

- 4 years, 11 months 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 :)

- 4 years, 11 months 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...

- 4 years, 11 months ago

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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.

- 4 years, 11 months ago

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That's what I was wondering!

- 4 years, 11 months ago

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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.

- 4 years, 11 months ago

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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..

- 4 years, 11 months ago

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\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}

- 4 years, 11 months ago

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$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=3⇒\frac{1}{1+a}=(1-\frac{1}{1+b})+(1-\frac{1}{1+c})+(1-\frac{1}{1+d})=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}$

Applying AM-HM inequality:

$\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\frac{1}{1+a}≥3\sqrt[3]{\frac{bcd}{(b+1)(c+1)(d+1)}}⇒\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}≥\frac{81abcd}{(a+1)(b+1)(c+1)(d+1)}⇒ abcd≤\frac{1}{81}$

- 4 years, 11 months 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.

- 4 years, 11 months 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, $$abcd = (\frac{1}{3}) ^ {4}$$

$$abcd = \frac{1}{81}$$

- 4 years, 11 months ago

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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.

- 4 years, 11 months ago

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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 Diego proposed.

The 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 $$F$$ does achieve 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.

Staff - 4 years, 11 months ago

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I like the idea of inverting the $$\frac{1}{1+t}$$ to $$\frac{1}{t} - 1$$ and then using a sum of logs rather than the original product. If I understand correctly, you don't even need to compute the derivatives since $$x,y,z,w$$ will all be the same solution of the same equation, $$f'(t) = k$$, where $$k$$ is the proportionality factor, and then we can solve for $$a,b,c,d$$ at the maximum using the original linear constraint and $$a=b=c=d$$.

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?

- 4 years, 11 months ago

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Well, I was actually a bit hasty in my claim. It is true that $$x, y, z, w$$ will all be solutions of the 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 the same solution of this equation.

However, if you look at both the first and the second derivatives, 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 immediately apply 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 $$(1/2,1)$$.

Staff - 4 years, 11 months ago

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