Let $a,b,c>0$. Prove that $$\frac{1+b}{1+c}a+\frac{1+c}{1+a}b+\frac{1+a}{1+b}c\geq a+b+c$$

Nice problem! This is equivalent with: $a(a+1)^2+b(b+1)^2>=(a+b)(a+1)(b+1)$. Consider the convex function $f(x)=x(x+1)^2$ We can write Jensen: $\frac{f(a)+f(b)}{2}\ge f\left( \frac{a+b}{2} \right)$. So it is enought to prove that $2f\left( \frac{a+b}{2} \right)\ge (a+b)(a+1)(b+1)\Leftrightarrow {{\left( \frac{a+b}{2}+1 \right)}^{2}}\ge (a+1)(b+1)$ and this is true, AM-GM

Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$

Since $$\frac{a(a+1)}{b+1}-a+\frac{b(b+1)}{a+1}-b = \frac{a(a-b)}{b+1} + \frac{b(b-a)}{a+1} = (a-b) \left( \frac{a}{b+1} - \frac{b}{a+1} \right) = (a-b) \frac{a^2+a-b^2-b}{(a+1)(b+1)} = (a-b)^2 \frac{a+b+1}{(a+1)(b+1)} \geq 0,$$it follows that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}\geq a+b.$$

Multiply both sides by $(a+1)(b+1)$ and get the inequality is equivalent to \[a(a+1)^2+b(b+1)^2\ge (a+b)(a+1)(b+1) \] Let $f(x)=x(x+1)^2$, which is convex. So we have \[f(a)+f(b)\ge 2f\left(\frac{a+b}{2}\right)=2\cdot \frac{a+b}{2}\cdot \left(\frac{a+b+2}{2}\right)^2=(a+b)\frac{(a+b+2)^2}{4}.\] Also, $(a+b+2)^2=a^2+b^2+2ab+4a+4b+4\ge 4ab+4a+4b+4=4(a+1)(b+1)$. So $(a+b)\frac{(a+b+2)^2}{4}\ge (a+b)(a+1)(b+1)$, as desired.

dangerousliri wrote: Show that for any positive real numbers $a$ and $b$ the following inequality hold, $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}\geq a+b.$$ My solution: $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}-(a+b)=(a-b)\left(\frac{a}{b+1}-\frac{b}{a+1}\right)=\frac{(a-b)^2(a+b+1)}{(a+1)(b+1)}\geq 0$$

Let $a,b,c>0$.Prove that $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{a+1}\geq a^2+b^2$$

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$ In fact I had this version also.

Let $a,b,c>0$.Prove that $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{c+1}+\frac{c^2(c+1)}{a+1} \geq a^2+b^2+c^2$$

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{a+1}\geq a^2+b^2$$ We can use also the Jensen for the convex function $f(x)=x^2(x+1)^2$

Let me provide yet another Jensen proof, applied on $f(x)=x^2$. Note that the inequality is equivalent to \[ \lambda f(a+1) + \overline{\lambda}f(b+1)\ge (a+1)(b+1),\qquad\text{where}\qquad \lambda = 1-\overline{\lambda} = \frac{a}{a+b}. \]Applying Jensen's and Cauchy-Schwarz, we find \begin{align*} \lambda f(a+1) + \overline{\lambda}f(b+1)&\ge f\bigl(\lambda(a+1)+\overline{\lambda}(b+1)\bigr)\\ &=\left(1+\frac{a^2+b^2}{a+b}\right)^2\\ &\ge \left(1+\frac{a+b}{2}\right)^2 \\ &\ge (a+1)(b+1), \end{align*}where the third line uses $2(a^2+b^2)\ge (a+b)^2$ valid by CS; and the last line follows from simple inequality $\frac{(x+y)^2}{4}\ge xy$ applied on $x=a+1$ and $y=b+1$.

Very nice! Congratulations!

We have \begin{align*} \frac{a(a + 1)}{b + 1} + \frac{b(b + 1)}{a + 1} &= \frac{a^2}{b + 1} + \frac{b^2}{a + 1} + \frac{a}{b + 1} + \frac{b}{a + 1} \\ &\ge \frac{(a + b)^2}{a + b + 2} + \frac{(a + b)^2}{2ab + a + b} \\ &\ge \frac{(a + b)^2}{a + b + 2} + \frac{2(a + b)^2}{(a + b)^2 + 2(a + b)} \\ &= \frac{(a + b)^2}{a + b + 2} \cdot \left( 1 + \frac{2}{a + b} \right) = a + b \end{align*}

If we multiply both sides by $(a+1)(b+1)$, we get $$a(a+1)^2+b(b+1)^2 \geq (a+b)(a+1)(b+1)$$ Then, $RHS-LHS$ gives $$(a-b)^2+a^3+b^3-a^2b-ab^2$$$(a-b)^2$ is always non-negative, so we need $a^3+b^3 \geq a^2b + ab^2$, which is equivalent to $$(a+b)(a^2-ab+b^2) \geq ab(a+b)$$ Because $a$ and $b$ are both positive reals, we can cancel out $a+b$ on both sides: $$a^2-ab+b^2 \geq ab$$$$a^2-2ab+b^2 \geq 0$$$$(a-b)^2 \geq 0$$ which is obviously true, so the inequality holds.

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$ Solution for this one? I believe Jensen doesn't work that nicely as in the previous posts (I tried splitting each fraction into two parts and applying Jensen (separately) for $f(x)=x^2$ and $g(x)=x(x+1)^2$ with weights of the type $\frac {a}{a+b+c}$ but it turned out to be disgusting) @below: Yup, you're right, that's probably the easiest solution.

Can we use the rearrangement inequality on the sequences $a(a+1), b(b+1), c(c+1)$ and $\frac{1}{a+1},\frac{1}{b+1},\frac{1}{c+1}$?

CT17 wrote: Can we use the rearrangement inequality on the sequences $a(a+1), b(b+1), c(c+1)$ and $\frac{1}{a+1},\frac{1}{b+1},\frac{1}{c+1}$? I was just writing that.

dangerousliri wrote: Show that for any positive real numbers $a$ and $b$ the following inequality hold, $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}\geq a+b.$$ Who proposed this problem?

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$ By Holder and C-S we obtain: $$\sum_{cyc}\frac{a^2+a}{b+1}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(ab+a)}+\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a)}=$$$$=\frac{a+b+c}{3\sum\limits_{cyc}(ab+a)}((a+b+c)^2+3(a+b+c))\geq\frac{a+b+c}{3\sum\limits_{cyc}(ab+a)}(3(ab+ac+bc)+3(a+b+c))=a+b+c.$$

Subtract $a + b$ from both sides to get\[[a(a+1) - b(b+1)]\left(\frac{1}{b+1} - \frac{1}{a+1} \right) \geq 0.\]Check that if $a > b$, then this is positive, and that if $a < b$, this is also positive. If $a = b$, then equality occurs. We are done.

Let $a,b,c>0$.Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$$$a^2+b^2+c^2 \geq\frac{ab(a+1)}{b+1}+\frac{bc(b+1)}{c+1}+\frac{ca(c+1)}{a+1}$$$$ab+bc+ca\geq\frac{ab(c+1)}{a+1}+\frac{bc(a+1)}{b+1}+\frac{ca(b+1)}{c+1}$$$$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{c+1}+\frac{c^2(c+1)}{a+1} \geq a^2+b^2+c^2 $$https://artofproblemsolving.com/community/c4h1583160p11605928 https://artofproblemsolving.com/community/c4h1583160p11625248 Let $a,b,c$ be non-negative numbers . Prove that$$\frac{a(bc+c+1)}{ca+a+1}+\frac{b(ca+a+1)}{ab+b+1}+\frac{c(ab+b+1)}{bc+c+1}\leq a+b+c.$$https://artofproblemsolving.com/community/c6h209651p3559606

Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}\geq \sqrt{2(a^2+b^2)}$$ sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{a+1}\geq a^2+b^2$$ Solution: $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{a+1}-(a^2+b^2)=(a-b)\left(\frac{a^2}{b+1}-\frac{b^2}{a+1}\right)=\frac{(a-b)^2(a^2+b^2+ab+a+b)}{(a+1)(b+1)}\geq 0$$

We know that $a^3+b^3\ge a^2b+b^2a$ and $a^2+b^2\ge 2ab$. $a^3+2a^2+a+b+b^3+2b^2\ge a^2b+ab^2+a^2+b^2+2ab+a+b \iff a(a+1)^2+b(b+1)^2\ge (a+b)(a+1)(b+1) \iff \frac{a^2+a}{b+1}+\frac{b^2+b}{a+1}\ge a+b$

Let $a,b $ be reals such that $ a+b\geq 1.$ Prove that $$\frac{a(a+1)}{b^2+a}+\frac{b(b+1)}{a^2+b} \geq 2$$

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a^2(a+1)}{b+1}+\frac{b^2(b+1)}{a+1}\geq a^2+b^2$$ dangerousliri wrote: Show that for any positive real numbers $a$ and $b$ the following inequality hold, $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{a+1}\geq a+b.$$

Attachments:

The conclusion $\Leftrightarrow \frac{a^2-ab}{b+1}+\frac{b^2-ab}{a+1}\geq 0$ $\Leftrightarrow \frac{(a-b)^2(a+b+1)}{(a+1)(b+1)}$ ,which is obvious.

sqing wrote: Let $a,b,c>0$.Prove that $$\frac{a(a+1)}{b+1}+\frac{b(b+1)}{c+1}+\frac{c(c+1)}{a+1} \geq a+b+c$$

Attachments:

sqing wrote: Let $a,b $ be reals such that $ a+b\geq 1.$ Prove that $$\frac{a(a+1)}{b^2+a}+\frac{b(b+1)}{a^2+b} \geq 2$$ Algebrical solution: $\frac{{{a}^{2}}+a}{{{b}^{2}}+a}-1=\frac{{{a}^{2}}-{{b}^{2}}}{{{b}^{2}}+a},\frac{{{b}^{2}}+b}{{{a}^{2}}+b}-1=-\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+b}$ $\frac{a(a+1)}{{{b}^{2}}+a}-1+\frac{b(b+1)}{{{a}^{2}}+b}-1=({{a}^{2}}-{{b}^{2}})\left( \frac{1}{{{b}^{2}}+a}-\frac{1}{{{a}^{2}}+b} \right)={{(a-b)}^{2}}(a+b)\frac{a+b-1}{({{b}^{2}}+a)({{a}^{2}}+b)}\ge 0$

Let $ a,b,c >0 $. Prove that$$\frac{a^2+1}{b+1}+ \frac{b^2+1}{c+1} + \frac{c^2+1}{a+1}\geq 2(\sqrt 2-1)(a+b+c)$$$$3.17325...(a^2+1) (b^2+1)(c^2+1)\geq (a+1)(b+1)(c+1)(a+b+c)$$https://artofproblemsolving.com/community/c6h3278811p30189127

$a(a+1)^2+b(b+1)^2 \geq (a+b)(a+1)(b+1)$ By Cauchy Schwarz $(a+b)(a(a+1)^2+b(b+1)^2) \geq (a(a+1)+b(b+1))^2$ So $LHS \geq \frac{(a(a+1)+b(b+1))^2}{a+b}$ Let's prove that $(a(a+1)+b(b+1))^2 \geq (a+b)^2(a+1)(b+1)$ $a(a+1)+b(b+1) \geq (a+b)\sqrt{(a+1)(b+1)}$ $(a^2+b^2)+(a+b) \geq (a+b)\sqrt{(a+1)(b+1)}$ By Cauchy Schwarz $\frac{(a+b)^2}{2}+(a+b) \geq (a+b)\sqrt{(a+1)(b+1)}$ $\frac{(a+b)}{2}+1 \geq \sqrt{(a+1)(b+1)}$ $\frac{(a+1)+(b+1)}{2} \geq \sqrt{(a+1)(b+1)}$ The last inequality is true because of AM-GM.

This problem guaranteed me the first place in $2022$ Back in $2022$ i didn't know what $AM-GM $ or $Cauchy$ were, I just knew that $(a-b)^2 \geq 0$ So i just multiplied everything and factorized and actually the given ineqaulity turns into: $(a-b)^2(a+b+1) \geq 0$ which is true

Let $a,b $ be reals such that $ a+b\geq 1.$ Prove that

$$ \frac{a(2a+1)}{b^2+a}+\frac{b(2b+1)}{a^2+b} \geq \frac{8}{3} $$#25