Note that $a/(2a+1) = \frac12 - \frac{1}{2(2a+1)}$. Likewise, $b/(3b+1)=\frac13 - \frac{1}{3(3b+1)}$ and $c/(6c+1) = \frac16 - \frac{1}{6(6c+1)}$. With this, it suffices to show \[ \frac{1}{4a+2}+\frac{1}{9b+3}+\frac{1}{36c+6}\ge \frac12. \]Next, write \[ \frac{1}{4a+2} = \frac{1}{12a+6}+\frac{1}{12a+6}+\frac{1}{12a+6}\qquad\text{and}\qquad \frac{1}{9b+3} = \frac{1}{18b+6}+\frac{1}{18b+6}. \]Using these expressions, we now apply AM-HM inequality in the following form: for any $n\in\mathbb{N}$ and positive $x_i$, $1\le i\le n$, we have $\textstyle \sum_{1\le i\le n}x_i^{-1}\ge n^2/(x_1+\cdots+x_n)$. With this, we deduce \[ \frac{1}{4a+2}+\frac{1}{9b+3}+\frac{1}{36c+6}\ge \frac{36}{36a + 36b+36c + 18+12+6} = \frac12, \]as requested.

Observe that ,$$x\to \frac{x}{nx+1}$$is concave in $[0,1]$ for all natural number $n$.Also equality holds at $a=\frac{1}{2},b=\frac{1}{3},c=\frac{1}{6}$.So it is natural to take tangents of the functions $f_n(x)=\frac{x}{nx+1}$ for $n=2,3,6$ at those points. We have $$f_n(x)=\frac{x}{nx+1}\le f'_n(\frac{1}{n})(x-\frac{1}{n})+f_n(\frac{1}{n})$$This is equivalent to , $$f_n(x)\le \frac{nx+1}{4n}$$. So we have, $$f_2(a)\le \frac{2a+1}{8},f_3(b)\le \frac{3b+1}{12},f_6(c)\le \frac{6c+1}{24}$$Adding them we get the required inequality.$\blacksquare$

Proposed by Karl Czakler. It is 2021 Austrian Federal Competition For Advanced Students, Part 1 p1. official source

sqing wrote: Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that$$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}.$$ Solution of Zhangyanzong : $$\frac{x}{x+1}\leq\frac{x+1}{4}(x\geq 0)$$$$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq\frac{1}{4}\left(a+\frac{1}{2}+b+\frac{1}{3}+c+\frac{1}{6}\right)= \frac{1}{2}.$$

Let $a,b\geq 0$ and $a+b=1.$ Find the maximum value of $\frac{a}{2a+1}+\frac{b}{3b+1}.$

sqing wrote: Let $a,b\geq 0$ and $a+b=1.$ Find the maximum value of $\frac{a}{2a+1}+\frac{b}{3b+1}.$ We will show $\frac{a}{2a+1}+\frac{b}{3b+1} \le \frac{5}{11}$. Multiply both sides by $-1$ and add $\frac{1}{2}+\frac{1}{3}$ to the both sides. $$\frac{1}{4a+2}+\frac{1}{9b+3} = \frac{9}{36a+18}+\frac{4}{36b+12} \ge \frac{(3+2)^2}{36+18+12} = \frac{25}{66}$$

sqing wrote: Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that$$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}.$$(Marian Dinca) $\frac{a}{2a+1}=\frac{1}{2}\cdot (1-\frac{1}{2a+1})$. We can change expressions with $b,c$ similarly. It is enough to prove that $$\frac{3}{2a+1}+\frac{2}{3b+1}+\frac{1}{6c+1}\ge 3.$$By C-S: $$(\frac{3}{2a+1}+\frac{2}{3b+1}+\frac{1}{6c+1})(3(2a+1)+2(3b+1)+6c+1)\ge6^2$$equality is reached when $a=1/2,b=1/3,c=1/6$. $\square$

sqing wrote: Let $a,b\geq 0$ and $a+b=1.$ Find the maximum value of $\frac{a}{2a+1}+\frac{b}{3b+1}.$ Let $f(a) = \frac{a}{2a+1} + \frac{a - 1}{3a - 4}$ which is our given expression. $f'(a) = 0$ $\implies$ (after some simplification) $\implies \frac{1}{(2a+1)^2} - \frac{1}{(3a - 4)^2}=0$ $\implies (3a - 4)^2 - (2a + 1)^2 = 0$ $\implies (a - 5)(5a - 3) = 0$ $\implies a = \frac{3}{5} $(abandoning a=5 then b=-4) So max $f(a) = f(\frac{3}{5}) = \frac{5}{11}$

Let $a,b\geq 0$ and $a+b=1.$ Prove that$$\frac{a}{2a+1}+\frac{b}{3b+1}\leq \frac{5}{11} .$$

sqing wrote: Let $a,b\geq 0$ and $a+b=1.$ Prove that$$\frac{a}{2a+1}+\frac{b}{3b+1}\leq \frac{5}{11} .$$ Solution of Zhangyanzong: $$\frac{a}{2a+1}+\frac{b}{3b+1}= \frac{5}{6} -\left( \frac{\frac{9}{4} }{9a+ \frac{9}{2} } + \frac{1}{9b+3} \right) \leq \frac{5}{6} -\frac{(\frac{3}{2}+1)^2}{9a+ \frac{9}{2}+9b+3 }= \frac{5}{11} .$$

$\text{One-variable inequality}\Rightarrow\text{Calculus}$ Let $f:[0,1]\to\mathbb R$ be a continuously differentiable function defined by $f(x)=\frac x{2x+1}+\frac{1-x}{4-3x}$. We want to find the maximum of $f$. Easily, $f'(x)=\frac{(5x-3)(x-5)}{(2x+1)^2(3x-4)^2}$, so we only need to consider $x=0.6$, which indeed gives the maximum of $f(0.6)=\frac5{11}$ as desired.

Let $a,\ b,\ c$ be positive real numbers such that $a+b+c=1.$ Prove that $$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}.$$ 2021 Austrian Mathematical Olympiad, proposed by Karl Czakler Solution by Kunihiko Chikaya, on May 26, 2021 $$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}\ \cdots (E)$$ $\frac{3}{2a+1}+\frac 34(2a+1)\geq 3\Longleftrightarrow \frac{3}{2a+1}\geq -\frac 32a+\frac 94.$ Analogously, $\frac{2}{3b+1}\geq -\frac 32b+\frac 32,\ \frac{1}{6c+1}\geq -\frac 32+\frac 34.$ Then, $\frac{3}{2a+1}+\frac{2}{3b+1}+\frac{1}{6c+1}\geq -\frac 32(a+b+c)+\frac 92=3.$ Thus, $6\times (E): 3-\frac{3}{2a+1}+2-\frac{2}{3b+1}+1-\frac{1}{6c+1}\leq 3.$ Equality : $2a=1,\ 3b=1,\ 6c=1\Longleftrightarrow a=\frac 12,\ b=\frac 13,\ c=\frac 16.$

Generalizaton of 2021 Austrian Mathematical Olympiad, proposed by Karl Czakler.

Let $a_1,a_2,\cdots,a_n ;k_1,k_2,\cdots,k_n (n\ge 2)$ be positive numbers such that $a_1+a_2+\cdots+a_n=S$ and $\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}=T.$ Prove that$$\frac{a_1}{k_1a_1+1}+\frac{a_2}{k_2a_2+1}+\cdots+\frac{a_n}{k_na_n+1}\leq\frac{S+T}{4}.$$ Let $a_1,a_2,\cdots,a_n ;k_1,k_2,\cdots,k_n (n\ge 2)$ be positive numbers such that $a_1+a_2+\cdots+a_n\leq S$ and $\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}\leq T.$ Prove that$$\frac{a_1}{k_1a_1+1}+\frac{a_2}{k_2a_2+1}+\cdots+\frac{a_n}{k_na_n+1}\leq\frac{S+T}{4}.$$It still is a very simple inequality.

sqing wrote: Let $a_1,a_2,\cdots,a_n ;k_1,k_2,\cdots,k_n (n\ge 2)$ be positive numbers such that $a_1+a_2+\cdots+a_n\leq S$ and $\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}\leq T.$ Prove that$$\frac{a_1}{k_1a_1+1}+\frac{a_2}{k_2a_2+1}+\cdots+\frac{a_n}{k_na_n+1}\leq\frac{S+T}{4}.$$It still is a very simple inequality. Obviously $4a_ik_i\le (a_ik_i+1)^2.$ Thus $\frac{a_i}{k_ia_i+1}\le \frac{a_ik_i+1}{4k_i}=\frac14(a_i+\frac{1}{k_i}).$ Adding up all and we have the desired inequality. It holds when $a_ik_i=1.$

Note that, by $MA-MH$, \begin{align*} \dfrac{a}{2a+1}+\dfrac{b}{3b+1}+\dfrac{c}{6c+1}&=\dfrac{1}{\frac{1}{\frac{1}{2}}+\frac{1}{a}}+\dfrac{1}{\frac{1}{\frac{1}{3}}+\frac{1}{b}}+\dfrac{1}{\frac{1}{\frac{1}{6}}+\frac{1}{c}}\\ &\le\dfrac{\frac{1}{2}+a}{4}+\dfrac{\frac{1}{3}+b}{4}+\dfrac{\frac{1}{6}+c}{4}\\ &=\dfrac{1}{2}\\ \end{align*} We are done. $\square$

Substitute $a=3x$, $b=2y$ and $c=z$. The Inequality is now equivalent to \[ 3 \cdot \frac{x}{6x+1} + 2 \cdot \frac{y}{6y+1}+1 \cdot \frac{z}{6z+1} \leq \frac{1}{2} \]Define $f(t)=\frac{t}{6t+1}$. Note that $f''(t)=\frac{-12}{(6t+1)^3} $ which implies that $f$ is concave for $t>0$. So by weighted Jensen: \[ 3 \cdot \frac{x}{6x+1} + 2 \cdot \frac{y}{6y+1}+1 \cdot \frac{z}{6z+1} \leq 6 \cdot \frac{\frac{x}{2}+\frac{y}{3}+\frac{z}{6}}{6\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6} \right)+1}=\frac{a+b+c}{a+b+c+1}=\frac{1}{2} \]and we are done.

Let $a,b\geq 0$ and $a+b=1.$ Prove that$$\frac{a}{a+1}+\frac{b}{2b+1}\leq \frac{3}{5} .$$Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that$$\frac{a}{a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}\leq \frac{2}{3}.$$

sqing wrote: Let $a,b\geq 0$ and $a+b=1.$ Prove that$$\frac{a}{a+1}+\frac{b}{2b+1}\leq \frac{3}{5} .$$ Let $$\frac a{a+1}+\frac b{2b+1}=\frac a{a+1}+\frac{1-a}{3-2a}=f(a).$$Then $f'(a)=\frac{(3a-2)(a-4)}{(a+1)^2(2a-3)^2}$, and since $a\ne4$ the only critical points are $a=0,\frac23,1$. Testing, we have $f(0)=\frac13$, $f\left(\frac23\right)=\frac35$, and $f(1)=\frac12$, so $\frac35$ is indeed the maximum.

Cauchy Shwarts inequality

If we defined $x=2a+1$, $y=3b+1$, and $z=6c+1$, we would need to show that \[ \frac{1}{2x}+\frac{1}{3y} + \frac{1}{6z} \ge \frac{1}{2} \]while $\frac{x}{2}+\frac{y}{3} + \frac{z}{6} = 2$. It's just an application of Cauchy-Schwarz inequality as \[ \left( \frac{x}{2}+\frac{y}{3} + \frac{z}{6} \right) \left( \frac{1}{2x}+\frac{1}{3y} + \frac{1}{6z} \right) \ge 1. \]

tangent line method