hello, the function $f(x,y,z)=\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}+(1-x)(1-y)(1-z)$ is strictly convex in each variable, hence it's maximum occurs only at the endpoint of $0\le x,y,z\le 1$. We find the expression has the value $1$ at all eight points. Sonnhard.

Im sorry but what does it mean to be "stricly convex at each variable" and how do we check for it?

Does anyone know of a manipulative solution to this that does not use convexity?

hello, see also here http://mathcircle.berkeley.edu/BMC4/Handouts/inequal/node4.html Sonnhard.

if $a+b+c=0$ then $a=0,b=0,c=0$ and the inequality is true. so we can assume that $a+b+c>0$ $1- \sum \frac{a}{b+c+1}\geq (1-a)(1-b)(1-c) \Longleftrightarrow$ $\sum( {\frac{a}{a+b+c}-\frac{a}{b+c+1}} ) \geq (1-a)(1-b)(1-c) \Longleftrightarrow$ $\sum \frac{a(1-a)}{(b+c+1)} \geq (1-a)(1-b)(1-c) (\sum a) \Longleftrightarrow$ $\sum {(a(1-a) (\frac{1}{b+c+1}- (1-b)(1-c)))} \geq 0$ it's sufficient to prove that $\frac{1}{b+c+1} \geq (1-b)(1-c)$ . define $x=1-b,y=1-c$ we should prove that $1 \geq xy(3-x-y)$ which it true because AM-GM : $ \frac {x+y+(3-x-y)}{3} \geq \sqrt[3] {xy(3-x-y)}$

Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}+(1-a)(1-b)(1-c) \geq 1,\] \[\frac{a^2}{b^2+1}+\frac{b^2}{c^2+1}+\frac{c^2}{a^2+1}+(1-a)(1-b)(1-c) \leq \frac{3}{2}.\]

As the double derivative is positive, it is $concave$ $down$. Also note that it is bounded. So by the $ sturm's$ $ priciple$(see link) aka $smoothing${https://www.tjhsst.edu/~2011akessler/lectures/tjusamo2011-inequalities.pdf} we know that the function attains its maximum at the end points. So {0,1} are points of local maxima and hence the inequality is proved by satisfying the end points into it. QED

aayush-srivastava wrote: As the double derivative is positive, it is $concave$ $down$. Also note that it is bounded. So by the $ sturm's$ $ priciple$(see link) aka $smoothing${https://www.tjhsst.edu/~2011akessler/lectures/tjusamo2011-inequalities.pdf} we know that the function attains its maximum at the end points. So {0,1} are points of local maxima and hence the inequality is proved by satisfying the end points into it. QED Your solution is correct. But if you had read the other posts (especially #2) you might have noticed that the very same solution has been there for more than 4 years now. (This is even more severe as you just revived a topic which has been inactive for more than a year...)

Dr Sonnhard Graubner wrote: hello, the function $f(x,y,z)=\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}+(1-x)(1-y)(1-z)$ is strictly convex in each variable, hence it's maximum occurs only at the endpoint of $0\le x,y,z\le 1$. We find the expression has the value $1$ at all eight points. Sonnhard. Yes but in this post the theorem hasn't been cited via which he claims that it attains its maximum at the end points.. I was just adding that

Mrdavid445 wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\] Someone proved that: For numbers $a,b,c$ in the interval $[0,1]$, $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \ge \frac{7}{8}.$

It's known!

Tintarn wrote: Your solution is correct. But if you had read the other posts (especially #2) you might have noticed that the very same solution has been there for more than 4 years now. (This is even more severe as you just revived a topic which has been inactive for more than a year...) Just for the record --- my understanding is that this is OK on the olympiad forums. It's not uncommon for olympiad students to solve, say, an old USAMO/ISL/etc. problem then find the corresponding thread and post their solution, even if it's similar to one already posted. (I do this all the time myself.) It can be instructive for both the poster and later forum readers. From here: darij grinberg wrote: Here in the advanced section, it is always ok to post solutions, even if they happen to be equivalent to solutions already posted (it can always happen that the new post is easier to understand than the older one, and there are other reasons as well). It is also ok to post hints if no solutions were posted. But if someone posted a complete proof of an inequality using Cauchy-Schwarz, then it is not ok anymore to post the one-liner "Cauchy-Schwarz pwns it".

Mrdavid445 wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\] We have \[1-(1-a)(1-b)(1-c)-\sum{\frac{a}{b+c+1}}=\frac{a(ab^3+ac^3+b)(1-a)+b(a^3b+bc+c)(1-b)+c(2b^3+ac)(1-c)}{(b+c+1)(c+a+1)(a+b+1)}\]\[+\frac{ab^2c(1-a)^2+bc^2(1-b)^2+ca(1-a)(1-b)^2}{(b+c+1)(c+a+1)(a+b+1)}\]\[+\frac{(2a^2b^2+ac^3+ab^2+2abc+b^2c)(1-a)(1-b)+a(2abc+b^2c+2ab+2ac+bc)(1-b)(1-c)+(a^2bc+a^2c^2+ab^3+b^2c^2)(1-c)(1-a)}{(b+c+1)(c+a+1)(a+b+1)}\ge{0}\]

Let $a,b,c{\in}(0,1]$ .Prove that$$\frac{1}{\sqrt{a+b+c}} \geq \frac{1}{\sqrt{3}} + (1-a)(1-b)(1-c) .$$

Mrdavid445 wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\] let\[a \le b \le c\]we just need to prove \[\frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\]which is \[(1-c)((1-a)(1-b)-\frac{1}{a+b+1}) \le 0.\]which is obvious as \[(1-a)(1-b) \le\frac{1}{(1+a)(1+b)} \le\frac{1}{a+b+1}\]

Grotex wrote: Mrdavid445 wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\] Someone proved that: For numbers $a,b,c$ in the interval $[0,1]$, $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \ge \frac{7}{8}.$ For numbers $a,b$ in the interval $[0,1]$, have$$1\geq\frac{a}{b+1}+\frac{b}{a+1}+(1-a)(1-b) \geq \frac{11-5\sqrt{5}}{2}.$$ Let $a,b,c,d$ in the interval $[0,1]$.Find theminimum value of $\frac{a}{b+c+d+1}+\frac{b}{c+d+a+1}+\frac{c}{d+a+b+1}+\frac{d}{a+b+c+1}+(1-a)(1-b)(1-c)(1-d) .$

g.c.boxd wrote: Im sorry but what does it mean to be "stricly convex at each variable" and how do we check for it? We double differentiate with respect to each of the variables and check if the double derivative (with respect to each variable) is positive or non-negative or non-positive or negative.

arqady wrote: It's known! Can you show some proof please ? It seems seen somewhere but can't find it !

Mrdavid445 wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \le 1.\] Let $a\geq b\geq c$ ,so $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq\frac{a+b+c}{b+c+1}+(1-a)(1-b)(1-c)=1-(1-a)\left(\frac{1}{b+c+1}-(1-b)(1-c)\right)\le 1.$ Generalization Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a^n}{b+c+1}+\frac{b^n}{c+a+1}+\frac{c^n}{a+b+1}+(1-a^n)(1-b^n)(1-c^n) \le 1.\]Where $n\geq 1.$

sqing wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a^n}{b+c+1}+\frac{b^n}{c+a+1}+\frac{c^n}{a+b+1}+(1-a^n)(1-b^n)(1-c^n) \le 1.\]Where $n\in N^+$ Just substitute $a=x^{1/n}$, $b=y^{1/n}$, and $c=z^{1/n}$. The resulting LHS is then stricly convex in $x,y,z$, and the old argument goes through as before.

test20 wrote: sqing wrote: Prove that for numbers $a,b,c$ in the interval $[0,1]$, \[\frac{a^n}{b+c+1}+\frac{b^n}{c+a+1}+\frac{c^n}{a+b+1}+(1-a^n)(1-b^n)(1-c^n) \le 1.\]Where $n\in N^+$ Just substitute $a=x^{1/n}$, $b=y^{1/n}$, and $c=z^{1/n}$. The resulting LHS is then stricly convex in $x,y,z$, and the old argument goes through as before. Are you sure? $$\frac{a^n}{b+c+1}+\frac{b^n}{c+a+1}+\frac{c^n}{a+b+1}+(1-a^n)(1-b^n)(1-c^n) \le \frac{a^n}{b^n+c^n+1}+\frac{b^n}{c^n+a^n+1}+\frac{c^n}{a^n+b^n+1}+(1-a^n)(1-b^n)(1-c^n) \le 1.$$

Prove or disprove that for numbers $a,b,c$ in the interval $[0,1]$, $$\frac{a}{b^2+c^2+1}+\frac{b}{c^2+a^2+1}+\frac{c}{a^2+b^2+1}+(1-a)(1-b)(1-c) \le 1.$$\[\frac{a^{\frac{1}{n}}}{b+c+1}+\frac{b^{\frac{1}{n}}}{c+a+1}+\frac{c^{\frac{1}{n}}}{a+b+1}+(1-a^{\frac{1}{n}})(1-b^{\frac{1}{n}})(1-c^{\frac{1}{n}}) \le 1.\]Where $n\in N^+$

sqing wrote: Let $a,b,c{\in}(0,1]$ .Prove that$$\frac{1}{\sqrt{a+b+c}} \geq \frac{1}{\sqrt{3}} + (1-a)(1-b)(1-c) .$$ Let $a,b,c{\in}(0,1]$ .Prove that$$\frac{1}{a+b+c}\geq \frac{1}{3} +\frac{9}{4} (1-a)(1-b)(1-c) .$$here

sqing wrote: Let $a,b,c{\in}(0,1]$ .Prove that$$\frac{1}{\sqrt{a+b+c}} \geq \frac{1}{\sqrt{3}} + (1-a)(1-b)(1-c) .$$ Let $a_1,a_2,\cdots,a_n {\in}(0,1]$ $ (n\ge 2).$ Prove that $$\frac{1}{\sqrt{a_1+a_2+\cdots+a_n}} \geq \frac{1}{\sqrt{n}} + (1-a_1)(1-a_2)\cdots (1-a_n).$$

sqing wrote: sqing wrote: Let $a,b,c{\in}(0,1]$ .Prove that$$\frac{1}{\sqrt{a+b+c}} \geq \frac{1}{\sqrt{3}} + (1-a)(1-b)(1-c) .$$ Let $a_1,a_2,\cdots,a_n {\in}(0,1]$ $ (n\ge 2).$ Prove that $$\frac{1}{\sqrt{a_1+a_2+\cdots+a_n}} \geq \frac{1}{\sqrt{n}} + (1-a_1)(1-a_2)\cdots (1-a_n).$$ Convexity method once again

Let $a,b,c$ are positive real numbers such that $abc=1 .$ Prove that$$a(a-1)+b(b-1)+c(c-1)+\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1.$$Let $a_1, ..., a_n\in (0,1).$ Prove that :$$\sum_{i=1}^{n} \frac{a_i}{S-a_i+1} + \prod_{i=1}^{n}( 1-a_i) \le 1$$Where $S=a_1+a_2+\cdots+a_n.$ New Zealand 2019

Who'd thought something new may be invented in this topic? Romania 2008 wrote: Let $ a,b\in[0;1]$. Prove the inequality: $$1-\dfrac{a+b}{2}+\dfrac{ab}{3}\geq\dfrac{1}{1+a+b}$$ Proof. Dr Sonnhard Graubner wrote: hello, the given inequality is equivalent to $ 3(1-a)(1-b)(a+b)+ab(1-a+1-b) \ge 0$, which is true because of $ 0\le a\le1$ and $ 0\le b\le1$. Sonnhard. Back to USAMO $$\sum_{cyc}\dfrac{c}{1+a+b}\le\sum_{cyc}\left(c-\dfrac{ac+bc}{2}+\dfrac{abc}{3}\right)=(a+b+c)-(ab+bc+ca)+abc=-(1-a)(1-b)(1-c)+1$$Thank you very much.

Prove that for numbers $a,b,c$ in the interval $[0,1]$, $$1\le\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}-(1-a)(1-b)(1-c) \le 2$$$$1\le\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}-2(1-a)(1-b)(1-c) \le 2$$$$1\le\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}+(1-a)(1-b)(1-c) \le \frac{3}{2}$$$$ \frac{1}{2} \le\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}+(1-a)(1-b)(1-c) \le 1$$$$\frac{1}{2}\leq \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+(1-a)(1-b)(1-c) \leq 1$$$$\frac{1}{k+1}\leq \frac{a}{a+k}+\frac{b}{b+k}+\frac{c}{c+k}+(1-a)(1-b)(1-c) \leq 1$$Where $k>0.$

Consider $f(a, b, c) = \frac{a}{b + c + 1} + \frac{b}{c + a + 1} + \frac{c}{b + c + 1} + (1 - a)(1 - b)(1 - c)$. Then we have that \begin{align*} \frac{\partial^2f}{\partial a^2} &= \frac{\partial}{\partial a} \left(\frac{1}{b + c + 1} - \left(\frac{b}{(c + a + 1)^2} + \frac{c}{(a + b + 1)^2} \right) + (b - 1)(1 - c) \right) \\ &= \frac{2b}{(c + a + 1)^2} + \frac{2c}{(a + b + 1)^2} \\ &\geq 0 \text{ over } [0, 1] \end{align*}Hence we have that the $f$ attains it's maximum for $a, b, c = 0, 1$. Now checking that $f \leq 1$ for all $a, b, c = 0, 1$ is trivial, and we are done. $\blacksquare$