Prove that for positive reals $a,b,c$ satisfying $a+b+c=3$ the following inequality holds: $$ \frac{1+a}{1+2b^3}+\frac{1+b}{1+2c^3}+\frac{1+c}{1+2a^3} \ge 2 $$$$ \frac{1+ca}{1+2b^3}+\frac{1+ab}{1+2c^3}+\frac{1+bc}{1+2a^3} \ge 2 $$Let $a,$ $b$ are non - negative real numbers and $a+b=2.$ Prove that $$1\leq\frac{a}{b^3+1}+\frac{b}{a^3+1}\leq2.$$Let $a,$ $b,$ $c$ are non - negative real numbers and $a+b+c=3.$ Prove that $$\frac{a}{2a^2+1}+\frac{b}{2b^2+1}+\frac{c}{2c^2+1}\leq1.$$Let $a,b,c>0$ such that $a+b+c=3$. Prove that: \[\frac{1}{2a^3+5}+\frac{1}{2b^3+5}+\frac{1}{2c^3+5} \le \frac{3}{7} \] Kimchiks926 wrote: Prove that for positive reals $a,b,c$ satisfying $a+b+c=3$ the following inequality holds: $$ \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \ge 1 $$

Kimchiks926 wrote: Prove that for positive reals $a,b,c$ satisfying $a+b+c=3$ the following inequality holds: $$ \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \ge 1 $$ $$(a-1)^2(a)\left(a+\frac 12 \right) \ge 0 \implies 1 \ge (1+2a^3)\left(1-\frac {2a}{3}\right) \implies \frac {1}{1+2a^3} \ge 1-\frac {2a}{3}$$similarly we have $\frac {1}{1+2b^3} \ge 1-\frac {2b}{3}$ and $\frac {1}{1+2c^3} \ge 1-\frac {2c}{3}$ thus $$\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \ge \sum \left(a-\frac {2ab}{3}\right)=\sum a -\frac 23 \sum ab =3 -\frac 23 \sum ab$$thus we only need to prove $\sum ab \le 3$ which is obviously true since $$\sum ab \le \frac {(a+b+c)^2}{3}=3$$$Q.E.D$

Let $a,b,c$ are non - negative real numbers such that $a+b+c=3.$ Prove that $$1\le \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \le 3 $$Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be non-negative numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove or disprove $$1\le \frac{a_1}{1+(n-1)a^3_2}+\frac{a_2}{1+(n-1)a^3_3}+\cdots+\frac{a_{n-1}}{1+(n-1)a^3_n}+\frac{a_n}{1+(n-1)a^3_1}\leq n.$$ sqing wrote: Let $a,$ $b$ are non - negative real numbers and $a+b=2.$ Prove that $$1\leq\frac{a}{b^3+1}+\frac{b}{a^3+1}\leq2.$$

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Let $a,b,c>0$ and $a+b+c=3.$ Prove that $$\frac{a}{2b^3+c}+ \frac{b}{2c^3+a}+\frac{c}{2a^3+b} \ge 1$$Let $a,b,c>0.$ Prove that$$\frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1}+\frac{1}{\sqrt{3(a^2+b^2+c^2)}}\geq\frac{4}{3}$$$$\frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1}+\frac{a+b+c}{a^2+b^2+c^2}>\frac{3}{2}$$

We use the tangent line trick. Notice that $\frac 1{1+2b^3} \geq 1-\frac{2b}3$ for all $b \geq 0$. Thus $$\sum_{\mathrm{cyc}} \frac a{1+2b^3} \geq \sum_{\mathrm{cyc}} a\left(1-\frac{2b}3\right) = 3-\frac 23 (ab+bc+ca) \geq 1$$is evident as $ab+bc+ca \leq 3$.

The tangent line trick gives the inequality $$\frac{1}{1 + 2b^3} \ge -\frac{2b}{3} + 1 \iff b(b - 1)^2(4b + 2) \ge 0,$$which clearly holds for $0 < b < 3$. Therefore $$\sum_{\text{cyc}} \frac{a}{1 + 2b^3} \ge \frac{-2(ab + bc + ca)}{3} + (a + b + c) \ge -2 + 3 = 1,$$where the last inequality follows from $ab + bc + ca \le \tfrac{1}{3}(a + b + c)^2$.

Same solution with anyone else, post for storage. $\frac{1}{1+2x^3}\leq 1-\frac23x\Longleftrightarrow x(x+\frac12)(x-1)^2\geq0$ So $\sum\frac{a}{1+2b^3}\geq\sum a-\frac23\sum ab$, so we only need $\sum ab\leq3$. we have $\sum ab\leq\frac{(a+b+c)^2}{3}=3$, and we're done.

Let $a,$ $b$ are non - negative real numbers and $a+b=2.$ Prove that $$2\geq \frac{a}{b^3+1}+\frac{b}{a^3+1}+\frac{4}{9}ab\geq \frac{13}{9}$$$$\frac{21}{10}>\frac{a}{b^3+1}+\frac{b}{a^3+1}+\frac{1}{2}ab\geq \frac{3}{2}$$$$\frac{9}{4}>\frac{a}{b^3+1}+\frac{b}{a^3+1}+ab\geq 2$$

Kimchiks926 wrote:

Splendid. And original problem

Prove that for positive reals $a,b,c$ satisfying $a+b+c=3$ the following inequality holds: $$ \frac{a}{2b^3+c}+ \frac{b}{2c^3+a}+\frac{c}{2a^3+b} \ge 1 $$

sqing wrote: Let $a,$ $b$ are non - negative real numbers and $a+b=2.$ Prove that $$2\geq \frac{a}{b^3+1}+\frac{b}{a^3+1}+\frac{4}{9}ab\geq \frac{13}{9}$$

Notice \[\sum_{\text{cyc}} \frac{a}{1+2b^3} = \sum_{\text{cyc}} \frac{a+2ab^3-2ab^3}{1+2b^3} = \sum_{\text{cyc}} a - \sum_{\text{cyc}} \frac{2ab^3}{1+2b^3}.\] Thus, it suffices to prove \[\sum_{\text{cyc}} \frac{ab^3}{1+2b^3} \le 1.\] Then, notice that $1+2b^3 \ge 3b^2$ by AM-GM, so it actually suffices to prove \[ab+bc+ca \le 3\] which is trivial. $\square$

Consider the function $f(x) = \tfrac{1}{1+2x^3}$. By Tangent Line Trick, we have the following inequality. \begin{align*} f(x) = \dfrac{1}{1+2x^3} \geq -\dfrac{2x}{3}+1 \quad \iff \quad (x - 1)^2 x \left(x + \tfrac{1}{2}\right) \geq 0 \end{align*}Therefore, $\tfrac{1}{1+2x^3} \geq -\tfrac{2x}{3}+1$. Substituting, it suffices to prove \begin{align*} \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} = -\tfrac{2}{3}\left(ab+bc+ca\right) + \left(a+b+c\right) \geq 1. \end{align*}As $a+b+c = 3$, then we want $3 \geq ab + bc + ca$. This is equivalent to showing the following. \begin{align*} \left(a+b+c\right)^2 \geq 3\left(ab+bc+ca\right) \quad \iff \quad a^2+b^2+c^2 \geq ab+bc+ca, \end{align*}which is clearly true.

We wish to show $\sum_{\text{cyc}} \dfrac{ab^3}{1 + 2b^3} \le 1$ and $1 + 2b^3 \ge 3b^2 \iff (b - 1)^2 (2b + 1) \ge 0$ so it is sufficient to show $\sum_{\text{cyc}} ab \le 3$ which is true as $\left(\sum_{\text{cyc}}a \right)^2 \ge 3 \sum_{\text{cyc}} ab$ and we are finished.

TLT IS BANNED!!!!!!!!!!!!!!!!!!! \[\frac{1}{1+2a^3}\ge 1-\frac23 a\]\[\sum_{\text{cyc}}a\left(1-\frac23 b\right)=3-\frac23\sum_{\text{cyc}}ab\ge 1.\]