Find the largest real number $k$, such that for any positive real numbers $a,b$, $$(a+b)(ab+1)(b+1)\geq kab^2$$
Problem
Source: 2019 CSMO Grade 10 P1
Tags: algebra, Inequality, China
WypHxr
30.07.2019 12:53
We try to find three positive integers $m,n,l$, $$a+\frac{b}{m}+ \cdots +\frac{b}{m} \geq (m+1)\sqrt[m+1]{\frac{ab^m}{m^m}};$$$$\frac{ab}{n}+\cdots+\frac{ab}{n}+1 \geq (n+1)\sqrt[n+1]{\frac{a^nb^n}{n^n}};$$$$\frac{b}{l}+\cdots+\frac{b}{l}+1 \geq (l+1)\sqrt[l+1]{\frac{b^l}{l^l}}.$$We hope: $\dfrac{1}{m+1}+\dfrac{n}{n+1}=1$, $\dfrac{m}{m+1}+\dfrac{n}{n+1}+\dfrac{l}{l+1}=2$, select $m=n=l=2$, we can get, $$(a+b)(ab+1)(b+1) \geq \dfrac{27}{4}ab^2.$$Select $a=1$, $b=2$ tell us $\dfrac{27}{4}$ is the best constant.
WypHxr
30.07.2019 12:57
Two of my student is same as follows: Let $$f(a,b)=\dfrac{(a+b)(ab+1)(b+1)}{ab^2}=\dfrac{(a+b)(b+\dfrac{1}{a})(b+1)}{b^2},$$we try to find minimum value of $f$. Consider to fix $b$, $\dfrac{\alpha f}{\alpha a}=\dfrac{b+1}{b}(1-\dfrac{1}{a^2})$, we know $f$ is increasing when $a \geq 1$, and decreasing when $0<a<1$, so $f(a,b) \geq f(1,b)=b+3+\dfrac{3}{b}+\dfrac{1}{b^2}$, Derivative for $b$, we get $f(1,b) \geq f(1,2)=\dfrac{27}{4}$.
MarkBcc168
30.07.2019 13:17
Let $x=\sqrt{ab}$ and $y=\sqrt{\tfrac{b}{a}}$. Thus $a=\tfrac{x}{y}$ and $b=xy$. The inequality becomes a symmetric inequality $$(x^2+1)(y^2+1)(xy+1)\geqslant k(xy)^2$$We claim that the answer is $k=\boxed{\tfrac{27}{4}}$, archived at $x=y=\sqrt{2}$. To prove the bound, note by Cauchy-Schwarz inequality, $$(x^2+1)(y^2+1)\geqslant (xy+1)^2$$and by AM-GM inequality $$xy+1 = \frac{xy}{2}+\frac{xy}{2}+1\geqslant 3\sqrt[3]{\frac{(xy)^2}{4}}$$thus $$(x^2+1)(y^2+1)(xy+1) \geqslant (xy+1)^3\geqslant \frac{27}{4}(xy)^2$$as desired.
WypHxr
30.07.2019 13:18
MarkBcc168 wrote: Let $x=\sqrt{ab}$ and $y=\sqrt{\tfrac{b}{a}}$. Thus $a=\tfrac{x}{y}$ and $b=xy$. The inequality becomes a symmetric inequality $$(x^2+1)(y^2+1)(xy+1)\geqslant k(xy)^2$$We claim that the answer is $k=\boxed{\tfrac{27}{4}}$, archived at $x=y=\sqrt{2}$. To prove the bound, note by Cauchy-Schwarz inequality, $$(x^2+1)(y^2+1)\geqslant (xy+1)^2$$and by AM-GM inequality $$xy+1 = \frac{xy}{2}+\frac{xy}{2}+1\geqslant 3\sqrt[3]{\frac{(xy)^2}{4}}$$thus $$(x^2+1)(y^2+1)(xy+1) \geqslant (xy+1)^3\geqslant \frac{27}{4}(xy)^2$$as desired. nice. Are you student or teacher? Are you in Ji An now?
MarkBcc168
30.07.2019 13:59
I'm a student from Thailand. I got problems from my friend participating there. (Thailand sends some students to compete at CSMO every year.)
Devastator
30.07.2019 14:21
Hi! Here’s my solution
Note that the inequality is equivalent to
$$\frac{(a+b)(ab+1)(b+1)}{ab^2} \geq k$$So we just need to find the minimum value of the $LHS$
If we let $t=ab$ and multiply the numerator and denominator by $b$, the expression on the $LHS$ becomes
$$\frac{(b^2+t)(b+1)(t+1)}{b^2t}$$
Now, by AM-GM, we have the following inequalities
$$b^2+t = \frac{b^2}{2} + \frac{b^2}{2} + t \geq 3 \sqrt[3]{ \frac{b^4t}{4} }$$$$b+1 = \frac{b}{2} + \frac{b}{2} + 1 \geq 3 \sqrt[3]{ \frac{b^2}{4} }$$$$t+1 = \frac{t}{2} + \frac{t}{2} + 1 \geq 3 \sqrt[3]{ \frac{t^2}{4} }$$
Plugging these inequalities into the above expression will give us
$$\frac{(b^2+t)(b+1)(t+1)}{b^2t} \geq \frac{27}{4}$$And equality holds when $b=t=2$, or equivalently, $a=1, b=2$
Hence, the answer is $\frac{27}{4}$
*this can definitely be rewritten with $a$ instead of $t$ but using $t$ just made it look cleaner for me
So I used the method of undetermined coefficients. Let $x,y$ be two positive integers to be determined later
So we’d want to AM-GM each of the 3 binomial factors in the numerator and get the product of the nonconstant part to be $b^2t$
Now, the AM-GM of the first factor would look like this
$$b^2+t = \frac{b^2}{x} + \frac{b^2}{x} + \dots + \frac{b^2}{x}+ \frac{t}{y} + \frac{t}{y} + \dots + \frac{t}{y} \geq (x+y) \cdot \frac{ b^{ \frac{2x}{x+y} } c^{ \frac{y}{x+y} } } {x^{ \frac{x}{x+y} } y^{\frac{y}{x+y}}}$$
Now, since we want the product of the $b$’s and $t$’s in the numerator to become $b^2t$ after our AM-GM, then the AM-GM of $b+1$ and $t+1$ should give out an exponent of $\frac{2y}{x+y}$ and $\frac{x}{x+y}$ for $b$ and $t$ respectively
Now, we force those exponents to appear in the following manner as shown below
$$b+1 = \frac{b}{2y} + \frac{b}{2y} + \dots + \frac{b}{2y} + \frac{1}{x-y} + \frac{1}{x-y} + \dots + \frac{1}{x-y} \geq
(x+y) \cdot \frac{ b^{ \frac{2y}{x+y} } }{ (2y)^{2y}(x-y){x-y} }$$*note that the above part will only work if $x>y$, and thankfully, it does as we shall see later
$$t+1 = \frac{t}{x} + \frac{t}{x} + \dots + \frac{t}{x} + \frac{1}{y} + \frac{1}{y} + \dots + \frac{1}{y} \geq
(x+y) \cdot \frac{ c^{ \frac{x}{x+y} } }{ x^xy^y }$$
Now we’re done forcing the exponents of $b$ and $t$ in the numerator to be $2$ and $1$ after the AM-GM, meaning that they will cancel with the denominator, leaving only our desired constant, which as of now is some ugly $x$ and $y$ garb.
Now, the last part is forcing the equality case to be possible. So the equality case for AM-GM in each of our three applications gives us the following three equations to solve
$$ \frac{b^2}{x} = \frac{t}{y} $$$$ \frac{b}{2y} = \frac{1}{x-y}$$$$ \frac{t}{x} = \frac{1}{y}$$We get that $b=\frac{x}{y}$ from the first and third equation
Plugging this back into the second equation and simplifying, we get $(x-2y)(x+y)=0$
Since we want $x,y$ to be positive integers, clearly we must have $x-2y=0$ rather than $x+y=0$
Now, we can simply pick $x=2,y=1$ to get the above solution
WypHxr
30.07.2019 14:29
MarkBcc168 wrote: I'm a student from Thailand. I got problems from my friend participating there. (Thailand sends some students to compete at CSMO every year.) We welcome you. I wish your friends get good marks.
IndoMathXdZ
30.07.2019 16:51
My solution. Equality holds when $(a,b) = (1,2)$. Notice that by AM - GM Inequality, \[ ab +1 = \frac{1}{2} ab + \frac{1}{2} ab + 1 \ge 3 \sqrt[3]{\frac{1}{4}a^2 b^2 } \]\[ a + b = \frac{1}{2} b +\frac{1}{2} b + a \ge 3 \sqrt[3]{\frac{1}{4} ab^2} \]\[ b + 1 = \frac{1}{2} b + \frac{1}{2}b + 1 \ge 3 \sqrt[3]{\frac{1}{4} b^2 } \]Multiply all of them and we get the result.
lemon94
04.08.2019 18:22
Does anyone know the results of the contest?