Set $ f(t) = \frac {1}{1 + e^t}$ and note that $ f(t) + f( - t) = 1$ fairly trivially. Setting $ x = \log a, y = \log b, z = \log c, w = \log d$, the problem transforms into For real numbers $ x\geq y \geq z \geq w$ such that $ x + y + z + w = 0$, show that \[ \frac {f(x) + f(y) + f(z)}{3} \geq f\left(\frac {x + y + z}{3}\right) \] Noting that the case where all the numbers are equal is trivial, we can assume that $ w < 0 < x$. Now, if $ x,y,z \geq 0$, the result easily follows from Jensen's Inequality (as $ f(t)$ is convex on $ t\geq 0$). We may therefore assume $ z<0$ also. If $ y<0$ as well, then let us take $ p=-y, q=-z, r=-w$. By the symmetry of the $ f$ function, we have to show \[ \frac {f(x) + 2 - f(p) - f(q)}{3} \geq f\left(\frac {r}{3}\right)\] or \[ f(p+q+r) + 2\geq 3f\left(\frac {r}{3}\right)+f(p)+f(q)\] Let's transform again, setting $ g(x) = 2f(x) - 1$. This doesn't actually change the inequality, but we now have $ g(x) = 2f(x) - 1 = 1 - 2f( - x) = - g( - x)$. I.e. $ g$ is odd. Thus \[ \frac {g(x) + g(y) + g(z)}{3} \geq g\left(\frac {x + y + z}{3}\right) \] becomes, using the constraints and the fact that $ g$ is odd \[ g(x) + g(y) + g(z) + 3g(\frac {w}{3}) \geq 0. \]

Let $ f(x)=\frac{1}{1+e^x}$. Verify that $ f''(x)=\frac{e^x(e^x-1)}{(1+e^x)^3}\ge 0$. Hence $ f$ is convex in $ \mathbb R^+$. Therefore from Jensen's inequality we conclude that \[ f(\ln a)+f(\ln b)+f(\ln c)\ge 3\cdot f\left(\frac{\ln a+\ln b+\ln c}{3}\right)=3\cdot f\left(\ln(abc)^{\frac 13}\right)\\ \iff \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge \frac{3}{1+\sqrt[3]{abc}}\]

Form our condition abcd=1 we can see that variable d dont have place in inequality and we have to prove it for any a>=0 , b>=0 , c>=0 . When we will reduce the left part of inequality to a common denominator we will achive : (3+2a+2b+2c+ab+bc+ac) / (1+a+b+c+ab+bc+ac+abc) >= 3 / 1 + abc^(1/3) If we will simplify it we will achive : (abc)^(1/3) * (3+2a+2b+2c+ab+ac+bc) >= a+b+c+2ab+2bc+2ac+3abc Lets simplify again : (a+b+c)(2(abc)^(1/3)-1)+(ab+bc+ac)((abc)^(1/3)-2)+3((abc)^(1/3)-abc) >= 0 If we do the Cauchis theorem to (a+b+c) and (ab+bc+ac) we will achive : a+b+c >= 3*(abc)^(1/3) ab+bc+ac >= 3*(abc)^(2/3) Lets put it into our inequality : 6*(abc)^(2/3) - 3*(abc)^(1/3) + 3abc - 6*(abc)^(2/3) + 3*(abc)^(1/3) - 3abc >= 0 0 >= 0 We achived the right inequality so the starting inequality was also right. So we are done. p.s n^k means n to deegre k

- consider fuction $ f(x) = \frac {1}{1 + e^x}$ - $ f''(x) = \frac {e^x(e^x - 1)}{(1 + e^x)^3}$ and $ f''(x)\geq 0$ when $ e^x\geq 1$ or in other words when $ x\geq 0$. function $ f$ is therefore convex for positive real numbers. - we can conclude from the formulation of the problem that $ abc\geq 1$ and $ a\geq 1$ we have to deal with $ 3$ cases since we don't know the $ max(1,b)$ and $ min(1,c)$ and we need to consider all possibilities. *Case1. $ a\geq b\geq c\geq 1$ - denote $ a = e^y$,, $ b = e^z$,, $ c = e^w$. where $ y,z,w$ are some positive reals due to the definition of $ a,b,c$ in this case. we have: $ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c} = f(y) + f(z) + f(w)$ - and by Jensen Inequality for convex functions: $ f(y) + f(z) + f(w)\geq 3f(\frac {y + z + w}{3})$ - now we also have: $ 3f(\frac {y + z + w}{3}) = \frac {3}{1 + e^{\frac {y + z + w}{3}}} = \frac {3}{1 + \sqrt [3]{e^y\cdot e^z\cdot e^w}} = \frac {3}{1 + \sqrt [3]{abc}}$ - therefore the inequalitiy holds *Case2. $ a\geq b\geq 1\geq c$ - denote $ c = \frac {1}{p}$. therefore $ p\geq 1$ and $ ab\geq p$ - the original inequality is therefore equivalent to: $ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {p}{1 + p}\geq \frac {3}{1 + \sqrt [3]{\frac {ab}{p}}}$ and further equivalent to (since $ 1 = \frac {1}{x + 1} + \frac {x}{x + 1}$): $ \frac {1}{1 + a} + \frac {1}{1 + b} + 1 \geq \frac {3}{1 + \sqrt [3]{\frac {ab}{p}}} + \frac {1}{1 + p}$ - now denote $ a = e^y$,, $ b = e^z$,, $ p = e^w$. where $ y,z,w$ are again positive reals and $ y\geq z$ and $ y + z\geq w$ due to the choice of $ a,b,c$ and the definition of $ p$ in this case. note also that $ 1 = \frac {1}{1 + e^0} + \frac {1}{1 + e^0}$ - the inequalitiy is therefore also equivalent to: $ f(y) + f(z) + 2f(0)\geq 3f(\frac {y + z - w}{3}) + f(w)$ where $ y,z,w,y + z - w\geq 0$. this last inquality holds due to the Karamata Inequality for convex functions since: $ y\geq \frac {y + z - w}{3}$ $ y + z\geq \frac {2(y + z - w)}{3}$ $ y + z + 0\geq \frac {3(y + z - w)}{3}$ and $ y + z + 0 + 0 = \frac {3(y + z - w)}{3} + w$ *Case3. $ a\geq 1\geq b\geq c$ - denote $ c = \frac {1}{p}$ and $ b = \frac {1}{q}$ where $ p,q\geq 1$ and $ a\geq pq$ - the original inequality is therefore equivalent to: $ \frac {1}{1 + a} + \frac {q}{1 + q} + \frac {p}{1 + p} \geq \frac {3}{1 + \sqrt [3]{\frac {a}{pq}}}$ and further equivalent to (since $ 1 = \frac {1}{x + 1} + \frac {x}{x + 1}$): $ \frac {1}{1 + a} + 2 \geq \frac {3}{1 + \sqrt [3]{\frac {a}{pq}}} + \frac {1}{1 + q} + \frac {1}{1 + p}$ - now denote $ a = e^y$,, $ q = e^z$,, $ p = e^w$ where $ y,z,w$ are again positive reals and $ y\geq z + w$ due to the choice of $ a,b,c$ and the definition of $ p,q$ in this case. note also that $ 2 = \frac {4}{1 + e^0}$ - the inequalitiy is therefore also equivalent to: $ f(y) + 4f(0)\geq 3f(\frac {y - z - w}{3}) + f(z) + f(w)$ where $ y,z,w,y - z - w\geq 0$. this last inquality holds due to the Karamata Inequality for convex functions since: $ y\geq \frac {y - z - w}{3}$ $ y + 0\geq \frac {2(y - z - w)}{3}$ $ y + 0 + 0\geq \frac {3(y - z - w)}{3}$ $ y + 0 + 0 + 0\geq \frac {3(y - z - w)}{3} + z$ and $ y + 0 + 0 + 0 + 0 = \frac {3(y - z - w)}{3} + z + w$ - all three possible cases are therefore prooven. therefore $ \frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c}\geq \frac {3}{1 + \sqrt [3]{abc}}$ for all $ a\geq b\geq c$ such that $ abc\geq1$,, ($ abcd = 1$) Q.E.D.

We attack this problem with convexity and symmetry. Let $ f$ be the real function $ v \mapsto \frac {1}{1 + \exp(v)}$. Let us denote $ x = \log a$, $ y = \log b$, $ z = \log c$, and $ w = \log d$. We know that $ x + y + z + w = 0$ and that $ x \ge y \ge z \ge w$; we wish to prove that \[ \frac {f(x) + f(y) + f(z)}{3} \ge f \left( \frac {x + y + z}{3} \right) . \] Lemma 1 (symmetry). For any real $ r$, $ f(r) + f( - r) = 1$. Proof. We note that \[ f(r) + f( - r) = \frac {1}{1 + e^r} + \frac {1}{1 + e^{ - r}} = \frac {e^{ - r/2}}{e^{ - r/2} + e^{r/2}} + \frac {e^{r/2}}{e^{r/2} + e^{ - r/2}} = 1. \qquad \blacksquare \] Lemma 2 (convexity). The function $ f$ is convex on the interval $ [0,\infty)$ and concave on the interval $ ( - \infty,0]$. Proof. The derivative of $ f$ is \[ f'(x) = - \frac {e^x}{(1 + e^x)^2} = \frac {1}{(1 + e^x)^2} - \frac {1}{1 + e^x}, \] so the second derivative of $ f$ is \[ f''(x) = e^x \left( \frac { - 2}{(1 + e^{x})^3} + \frac {1}{(1 + e^{x})^2} \right) = e^x \cdot \frac {e^x - 1}{(1 + e^x)^3} , \] which is evidently greater than or equal to zero on $ [0,\infty)$ and less than or equal to zero on $ ( - \infty,0]$. $ \blacksquare$ Lemma 3 (a bound). If $ r$ and $ s$ are reals such that $ r \ge \lvert s \rvert$, and $ \lambda_1, \lambda_2$ are positive reals such that $ \lambda_1 + \lambda_2 = 1$ and $ \lambda_1 r + \lambda_2 s \ge 0$, then \[ \lambda_1 f(r) + \lambda_2 f(s) \ge f(\lambda_1 r + \lambda_2 s) . \] Proof. If $ s \ge 0$, then by Lemma 2, we simply apply Jensen's Inequality and we are done. Hence we assume that $ s < 0$. Let $ (t,\rho(t))$ be the parametrization of the line in $ \mathbb{R}^2$ passing the points $ (r,f(r))$ and $ ( - r,f( - r))$, and let $ (t,\sigma(t))$ be the parametrization of the line passing through the points $ (r,f(r))$ and $ (s,f(s))$. By Lemma 1, the line $ (t,\rho(t))$ contains the point $ (0,1/2)$. Since $ s \in [ - r,0]$, it follows from the concavity of $ f$ on $ ( - \infty, 0]$ that \[ \sigma(s) = f(s) \ge \rho(s) . \] Since \[ \sigma(r) = f(r) = \rho(r), \] it follows that for any real $ t \in [s,r]$, $ \sigma(t) \ge \rho(t)$. In particular, \[ \lambda_1 f(r) + \lambda_2 f(s) = \sigma (\lambda_1 r + \lambda_2 s) \ge \rho (\lambda_1 r + \lambda_2 s) . \qquad (*) \] And since $ \lambda_1 r + \lambda_2 s \in [0,r]$, $ f$ is convex on that interval, and $ \rho$ is the linear function connecting the endpoints, \[ \rho(\lambda_1 r + \lambda_2 s) \ge f(\lambda_1 r + \lambda_2 s) . \qquad (**) \] We prove the lemma by combining $ (*)$ and $ (**)$. $ \blacksquare$ The problem will now follow from repeated application of Lemma 3. Now, $ x + y \ge \frac {x + y + z + w}{2} = 0$; since $ x\ge y$ and $ x + y \ge 0$, it follows that $ x \ge \lvert y \rvert$. Hence $ (r,s,\lambda_1,\lambda_2) = (x,y,1/2,1/2)$ satisfy the hypotheses of Lemma 3, so \[ f(x) + f(y) \ge 2 f \left( \frac {x + y}{2} \right) . \] Also, $ \frac {x + y + z}{3} \ge \frac {x + y + z + w}{4} = 0$. If $ \frac {x + y}{2} + z \le 0$, then $ \frac {x + y}{2} + w \ge 0$, so $ w \ge z$, a contradiction. Hence $ \frac {x + y}{2} + z \ge 0$. Therefore \[ (r,s,\lambda_1, \lambda_2) = \left( \frac {x + y}{2} , z, 2/3, 1/3 \right) \] satisfies the hypotheses of Lemma 3, so \[ \frac {2}{3} f \left( \frac {x + y}{2} \right) + \frac {1}{3} f(z) \ge f \left( \frac {x + y + z}{3} \right). \] In summary, \[ \frac {f(x) + f(y) + f(z)}{3} \ge \frac {2 f\bigl( (x + y)/2 \bigr) + f(z)}{3} \ge f \left( \frac {x + y + z}{3} \right), \] which is what we wanted to prove. $ \blacksquare$

Lemma: If $ n,m,p \in R^ +$, $ p\leq mn$.$ = > \frac {1}{p + m^2} + \frac {1}{p + n^2}\geq \frac {2}{p + mn}$ Dem: $ \frac {1}{p + m^2} + \frac {1}{p + n^2}\geq \frac {2}{p + mn}$ $ < = > (m^2 + n^2 + 2p)(mn + p)\geq 2(m^2n^2 + n^2p + m^2p + p^2)$ $ < = > (m - n)^2(nm - p)\geq 0$, $ Q.E.D.$ In the problem.$ a\geq b \geq c \geq d > 0$, and $ abcd = 1$...(1) if: $ a = \frac {x^4}{w}$, $ b = \frac {y^4}{w}$, $ c = \frac {z^4}{w}$ $ = > d = w^3$, where $ x,y,z,w \in R^ +$ In (1): $ xyz = 1$, $ x\geq y \geq z \geq w$..(2), From (1) and (2), $ 1 = abcd \geq d^4$ $ = > d \leq 1$, $ 1 = xyz\geq z^3$ $ = > z\leq 1$, but $ d = w^3$=> $ w\leq z \leq 1.$ ...(3). Replacing in the inequality. $ \frac 1{1 + a} + \frac 1{1 + b} + \frac 1{1 + c} \geq \frac {3}{1 + \sqrt [3]{abc}}.$ $ - >$ $ \frac 1{w + x^4} + \frac 1{w + y^4} + \frac 1{w + z^4} \geq \frac {3}{1 + w}$. now,how $ z\leq 1$ $ = > xyz \leq xy$ $ = > xy\geq 1$, $ = > x^2y^2\geq xy \geq 1 \geq w$ $ = > x^2y^2\geq w$, Applying Lemma to $ x^2,y^2,w$. $ = > \frac 1{w + x^4} + \frac 1{w + y^4}\geq \frac {2}{x^2y^2 + w}$, but $ xyz = 1$ $ = > \frac 1{w + x^4} + \frac 1{w + y^4}\geq \frac {2z^2}{1 + z^2w}$ $ \frac 1{w + x^4} + \frac 1{w + y^4} + \frac 1{w + z^4} \geq \frac {2z^2}{1 + z^2w} + \frac 1{w + z^4}$. Now. $ \frac {2z^2}{1 + z^2w} + \frac 1{w + z^4} \geq \frac {3}{1 + w}$...(4) $ < = > (2z^6 + 3z^2w + 1)(w + 1) \geq 3(z^4 + w + w^6 + z^2w^2)$ $ < = > (z - 1)^2(2z^4 + 4z^3 + 3z^2 + 2z + 1 - w(z^4 + 2z^3 + 3z^2 + 4z + 2))\geq 0$ $ < = > (2z^4 + 4z^3 + 3z^2 + 2z + 1 - w(z^4 + 2z^3 + 3z^2 + 4z + 2)) \geq 0$...(*) $ M = z^5 - z^3 + 2z^2 - 1 = (z - 1)(z^4 + z^3 + z + 1)\leq 0$ $ S = (2z^4 + 4z^3 + 3z^2 + 2z + 1 - w(z^4 + 2z^3 + 3z^2 + 4z + 2))$ $ S + M = (z - w)(z^4 + 2z^3 + 3z^2 + 4z + 2)\geq 0$ $ (z\geq w$ $ = > S + M\geq 0 \geq M$ $ = > S\geq 0$..=> (*) is true.=> (4) is true... $ = > \frac 1{w + x^4} + \frac 1{w + y^4} + \frac 1{w + z^4} \geq \frac {2z^2}{1 + z^2w} + \frac 1{w + z^4} \geq \frac {3}{1 + w}$ $ \frac 1{w + x^4} + \frac 1{w + y^4} + \frac 1{w + z^4} \geq \frac {3}{1 + w}$, $ Q.E.D.$

imo shortlist 1998 problem 2.n=3

I misunderstand something There are 4 variables but inequality only have 3 variables???

We put $ a = x^6$ , $ b = y^6$ , $ c = z^6$ and $ d = w^6$ From the condition $ x\geq y\geq z\geq w > 0$ and $ xyzw = 1$ and we easily obtain that $ xy\geq 1$ , $ xyz\geq 1$ and $ xyz^2\geq 1$ (1) (just replace at $ xyzw = 1$ the $ zw$ with $ xy$ , the $ w^3$ with $ xyz$ and the $ w$ with $ z$ ) We have that $ \frac {1}{1 + x^6} + \frac {1}{1 + y^6}\geq\frac {2}{1 + x^3y^3}$ which after the expanding is equivalent to $ (x - y)^2(xy - 1)(x^2 + xy + y^2)^2(1 + xy + x^2y^2)\geq 0$ which is clearly true since $ xy\geq 1$ So we have to prove that $ \frac {2}{1 + x^3y^3} + \frac {1}{1 + z^6}\geq\frac {3}{1 + x^2y^2z^2}$ which after expanding is equivalent to $ (xy - z^2)^2(2x^2y^2z^4 + x^3y^3z^2 - 2xy - z^2)\geq 0$ so we have only to prove that $ 2x^2y^2z^4 + x^3y^3z^2 - 2xy - z^2\geq 0$ . We have from (1) that $ x^3y^3z^2\geq x^2y^2$ and $ x^2y^2z^4\geq z^2$ and also $ x^2y^2z^4\geq 1$ . Adding these three we have that $ 2x^2y^2z^4 + x^3y^3z^2\geq x^2y^2 + 1 + z^2$ so we have to prove that $ x^2y^2 + 1 + z^2 - 2xy - z^2\geq 0$ which is equivalent to $ (xy - 1)^2\geq 0$ which is clearly true and we are done

We must prove that $ \frac 1{1+a} + \frac 1{1+b} + \frac 1{1+c} \geq \frac {3}{1+\sqrt[3]{abc}}$. Denote $ a=e^x,b=e^y,c=e^z$. Consider the fuction: $ f(t)=\frac 1{1+e^t}$, we have $ f"(t)=\frac{2e^t}{(1+e^t)^2}>0$. By Jensen inequality, we have $ f(x) + f(y) + f(z) \geq 3f(\frac{x+y+z}{3})$. Hence, the solution is complete. .

Since the harmonic mean is less than or equal to the geometric mean, let's switch the geometric mean in the denominator to a harmonic mean to make the right side larger. We find that the inequality still holds as follows, proving the desired result. Let $ P = abc$, $ S = a + b + c$, and $ S_2 = ab + bc + ac$. \[ \frac1{1 + a} + \frac1{1 + b} + \frac1{1 + c} \geq \frac3{1 + \frac3{\frac1a + \frac1b + \frac1c}} \] After some algebra (I'm short on time), this leads to \[ 9P + 6SP + 2PS_2 \geq 0 \] , which is clearly true.

Valentin Vornicu wrote: If $ a\geq b\geq c\geq d > 0$ such that $ abcd=1$, then prove that \[ \frac 1{1+a} + \frac 1{1+b} + \frac 1{1+c} \geq \frac {3}{1+\sqrt[3]{abc}}.\] https://artofproblemsolving.com/community/c6h433168p2447520 Let $a,b,c>0, \ \ abc\geq 1$. Prove that \[ \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{3}{\sqrt[3]{abc}+1}\]