For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$
Problem
Source: Chinese Girl's MO 2007
Tags: inequalities, China, CGMO
01.01.2008 11:11
Hmm...I have been spending a while on this problem. It is pretty easy to bash with lagrange multipliers. Of course, that is an unsatisfying solution. Directly using cauchy/jensen to reduce the square roots doesn't work. I think we need some random manipulation to solve this...
01.01.2008 15:30
We have another one. It is similar to above but it is a symmetric inequality. \[ \sum \sqrt{x+\frac{(y-z)^2}{12}}\le \sqrt{3}\] with $ x,y,z\ge 0$ and $ x+y+z=1$.
01.01.2008 18:30
So, please post solutions to both
02.01.2008 20:34
I think that we should make a substitution of (a,b,c)=(x^2,y^2,z^2). Then proceed to showing that y=z, but I don't know.
07.01.2008 21:39
Am I allowed to post solution to this inequality? Here's a completely elementary solution, using only AM-GM for two numbers: Assume $ b\ge c$. Let $ z=\frac{b+c}2$ and $ \Delta=\frac{b-c}2\ge0$. Note that $ \Delta^2\le\Delta\le z\le\frac12$. We'll use the inequality $ \boxed{\sqrt{z^2-\Delta^2}\le z-\Delta^2}$. To check that this one is true, square it, and then it becomes $ \Delta^2(2z-1)\le\Delta^4$, which is true since $ z\le\frac12$. Back to the problem. Using our notations, the inequality becomes: $ \sqrt{a+\Delta^2}+\sqrt{z+\Delta}+\sqrt{z-\Delta}\le\sqrt3$. After squaring and reducing some terms, keeping in mind that $ a+2z=1$, we get $ \Delta^2+2\sqrt{z^2-\Delta^2}+2\sqrt{a+\Delta^2}\left[\sqrt{z+\Delta}+\sqrt{z-\Delta}\right]\le2=2a+4z$. Using the inequality stated at the beginning, we see that $ \Delta^2+2\sqrt{z^2-\Delta^2}\le 2z-\Delta^2$. Hence, we are left with proving that $ -\frac{\Delta^2}2+2\sqrt{a+\Delta^2}\left[\frac{\sqrt{z+\Delta}+\sqrt{z-\Delta}}2\right]\le a+z$, or $ 2\sqrt{a+\Delta^2}\left[\frac{\sqrt{z+\Delta}+\sqrt{z-\Delta}}2\right]\le (a+\Delta^2)+\left(z-\dfrac{\Delta^2}2\right)$. We recognize AM-GM here; so, what's left is proving that $ \sqrt{z+\Delta}+\sqrt{z-\Delta}\le\sqrt{4z-2\Delta^2}$. After squaring this becomes again $ \sqrt{z^2-\Delta^2}\le z-\Delta^2$, proved at the beginning. Hope it is correct.
10.01.2008 12:30
Good work, freemind. I like the problem and your solution too.
11.04.2008 07:44
I have posted a solution here (with only am-gm ). http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=199502
22.04.2014 12:42
We have \[\frac{\sqrt b+\sqrt c}{\sqrt 2}=\sqrt{b+c-\frac{(\sqrt b-\sqrt c)^2}{2}}\le \sqrt{b+c-\frac{( b- c)^2}{4}}\] therefore from this and Cauchy-Schwarz it follows that \[\left( \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2\le \left(a+\frac{(b-c)^2}{4}+\frac{(\sqrt b+\sqrt c)^2}{ 2}\right)(1+2)\] \[\le 3\left(a+\frac{(b-c)^2}{4}+b+c -\frac{(b-c)^2}{4}\right)=3 \]
22.04.2014 14:01
I have a solution(《Middle school mathematics》(China Wuhan)No.4(2008)): $(\sqrt b+\sqrt c)^2\le 2(b+c)\le 2(a+b+c)=2\Rightarrow$ $(\sqrt b-\sqrt c)^2\left(2-(\sqrt b+\sqrt c)^2\right)\ge 0\iff$ $(\sqrt b-\sqrt c)^2\ge \frac{1}{2}( b- c)^2\Rightarrow$ $(\sqrt b-\sqrt c)^2+\left(\sqrt{3}(\sqrt b+\sqrt c)-2\right)^2\ge \frac{1}{2}( b- c)^2\iff$ $\left(\sqrt{3}-\sqrt b-\sqrt c \right)^2\ge a+\frac{1}{4}( b- c)^2\iff$ $ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}.$
22.04.2014 16:01
The following inequality is also true. For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that\[ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\geq 1.\]
22.04.2014 18:01
sqing wrote: The following inequality is also true. For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that\[ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\geq 1.\] $ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\geq\sqrt a+\sqrt{b}+\sqrt{c}\geq a+b+c\geq1$.
23.04.2014 02:37
For $ a,b,c\geq 0$ with $ a+b+c=3.$ Prove that\[ \sqrt{ab+\frac{(a-b)^2}{3}}+\sqrt{bc}+\sqrt{ca}\leq 3.\]
Attachments:

24.04.2014 20:36
let $t= \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}$,we must prove $t\leqslant\sqrt{3}$, hence,\begin{align*} &{\kern 9pt}3(a+b+c)+\frac34(b-c)^2\\ &=t^2+(\sqrt b-\sqrt c)^2+(\sqrt{a+\frac{(b-c)^2}{4}}-\sqrt b)^2+(\sqrt{a+\frac{(b-c)^2}{4}}-\sqrt c)^2\\ &\geqslant t^2+(\sqrt b-\sqrt c)^2+\frac12(\sqrt b-\sqrt c)^2\\ &=t^2+\frac32(\sqrt b-\sqrt c)^2\\ &\geqslant t^2+\frac34(b-c)^2\\ \end{align*} that is \[3(a+b+c)\geqslant t^2\Longrightarrow t\leqslant \sqrt{3}\]
08.05.2015 02:32
easternlatincup wrote: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$ Romania NMO 2015,9th grade,Problem 4: Let $a,b,c,d \ge 0$ real numbers so that $a+b+c+d=1$.Prove that $\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}} +\sqrt{b}+\sqrt{c}+\sqrt{d} \le 2.$
15.11.2015 04:14
easternlatincup wrote: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$ Strengthening: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that$$ \sqrt{a+\frac{155(b-c)^2}{456}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}.$$
06.03.2016 05:20
For $ a,b,c,d\geq 0$ with $ a+b+c+d=1$, prove that\[ \sqrt{a+\frac{2(b-c)^2}{9}+\frac{2(c-d)^2}{9}+\frac{2(d-b)^2}{9}}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 2.\](Proposed by Marius Stanean, Zalau, Romania)
14.03.2016 03:21
Generalization Of Some Famous Inequalities
14.03.2016 06:26
sqing wrote: Generalization Of Some Famous Inequalities Thx for your quotation. 谢谢宋老师关注我的成果…… Grotex(何嘉豪)
14.03.2016 06:35
Grotex wrote: sqing wrote: Generalization Of Some Famous Inequalities Thx for your quotation. 谢谢宋老师关注我的成果…… Grotex(何嘉豪) Amazing. Envy you young people. Good luck.
14.03.2016 15:42
sqing wrote: Amazing. Envy you young people. Good luck.
Yeah^-^ sqing(Mr. Song),I am fond of ineq and output some interesting one,but I am surprised that you know me well......
14.03.2016 16:21
Grotex wrote: sqing wrote: Amazing. Envy you young people. Good luck.
Yeah^-^ sqing(Mr. Song),I am fond of ineq and output some interesting one,but I am surprised that you know me well......
Given $a,b,c \ge 0$ such that $a+b+c=4$. Prove \[\sqrt{2a+\dfrac{(b-c)^2}{8}}+\sqrt{b}+\sqrt{c} \le 4\] $$(b-c)^{2}=\left(\sqrt{b}-\sqrt{c}\right)^{2}\left(\sqrt{b}+\sqrt{c}\right)^{2}\leq 2(b+c)\left(\sqrt{b}-\sqrt{c}\right)^{2}\leq 8\left(\sqrt{b}-\sqrt{c}\right)^{2}$$$\implies LHS\leq \sqrt{2a+\left(\sqrt{b}-\sqrt{c}\right)^{2}}+\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^{2}}\leq \sqrt{2\left[2a+\left(\sqrt{b}-\sqrt{c}\right)^{2}+\left(\sqrt{b}+\sqrt{c}\right)^{2}\right]}=2\sqrt{a+b+c}=4$ https://artofproblemsolving.com/community/c6h3009101p27028117
13.06.2017 12:05
20.11.2018 21:28
13.03.2019 10:47
sqing wrote: easternlatincup wrote: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$ Strengthening: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that$$ \sqrt{a+\frac{155(b-c)^2}{456}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}.$$ For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(b-c)^2}{3}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$$$ \sqrt{a+\frac{(b-c)^2}{6}}+\sqrt{a+\frac{(c-a)^2}{6}}+\sqrt{c}\leq \sqrt{3}$$$$ \sqrt{a+(1-\frac{\sqrt{3}}{2})(b-c)^{2}}+\sqrt{b+(1-\frac{\sqrt{3}}{2})(c-a)^{2}}+\sqrt{c+(1-\frac{\sqrt{3}}{2})(a-b)^{2}}\leq \sqrt{3}.$$p/6054957167
Attachments:


26.04.2019 17:28
For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(b-c)^2}{2}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$Proof: $$\sqrt{a+\frac{(b-c)^2}{2}}+\sqrt{b}+\sqrt{c}=\sqrt{a+\frac{(b-c)^2}{2}}+\sqrt{\frac{1}{4}(\sqrt{b}+\sqrt{c})^2}+\sqrt{\frac{1}{4}(\sqrt{b}+\sqrt{c})^2}$$$$\sqrt{3\left(a+\frac{(b-c)^2}{2}+\frac{1}{4}(\sqrt{b}+\sqrt{c})^2+\frac{1}{4}(\sqrt{b}+\sqrt{c})^2\right)} =\sqrt{3}.$$For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(b-c)^2}{12}}+\sqrt{b+1}+\sqrt{c+1}\leq3$$$$ \sqrt{2a+\frac{(b-c)^2}{2}}+\sqrt{b}+\sqrt{c}\leq 2$$$$ \sqrt{\frac{2}{3}a+\frac{(b-c)^2}{6}}+\sqrt{b}+\sqrt{c}\leq \frac{2}{3}\sqrt{6}$$Proof: ... (lixin)
24.12.2019 03:14
sqing wrote: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that
For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(\sqrt b-\sqrt c)^2}{2}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$$$ c\sqrt{a+\frac{(b-c)^2}{3}}+a\sqrt{b}+b\sqrt{c}\leq \frac{1}{\sqrt{3}}$$
25.12.2019 05:07
easternlatincup wrote: For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$ Abdek: $$\sqrt{c}+\sqrt{b} \le \sqrt{2(b+c)} \le\sqrt{ 2}\implies (\sqrt{b}-\sqrt{c})^2(2-(\sqrt{b}+\sqrt{c})^2)\ge 0\iff \frac{(\sqrt{b}-\sqrt{c})^2}{2}\ge \frac{(b-c)^2}{4}$$By Cauchy Shwarz inequality , $$\sqrt{a+\frac{(b-c)^2}{4}} +\sqrt{(\sqrt{b}+\sqrt{c})^2} \le\sqrt{a+\frac{(\sqrt{b}-\sqrt{c})^2}{2}} +\sqrt{(\sqrt{b}+\sqrt{c})^2} $$$$\le \sqrt{(1+2)\left(a+\frac{(\sqrt{b}-\sqrt{c})^2}{2}+\frac{(\sqrt{b}+\sqrt{c})^2}{2}\right)}=\sqrt{3}$$Equality occurs when $a=b=c=\frac{1}{3}.$ For $ a,b,c>0$ with $a+b+c=3.$ Prove that$$ \sqrt{ab+\frac{1}{3}(a-b)^2}+\sqrt{bc}+\sqrt{ca}\leq 3$$
Attachments:


25.06.2020 16:34
By Lagrange's Identity , we have$$(\sqrt b +\sqrt c) ^2 = 2(b+c)-(\sqrt b-\sqrt c)^2$$$$\implies \left(\frac{\sqrt b +\sqrt c}{\sqrt 2}\right)^2 = (b+c) -\left(\frac{\sqrt b -\sqrt c }{\sqrt 2}\right)^2 \leq (b+c)-\left(\frac{b-c}{2}\right)^2$$. Now we write $$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{a+\frac{(b-c)^2}{4}} + \sqrt 2 \cdot \sqrt{(b+c)-\frac{(b-c)^2}{4}} \overset{\text{C-S}}{\leq} \left(\sqrt{1+2}\right) \cdot \left(\sqrt{a+\frac{(b-c)^2}{4}+(b+c)-\frac{(b-c)^2}{4}}\right) \leq \sqrt 3 \cdot \sqrt{a+b+c} = \sqrt 3 $$. We're done . Equality occurs when $a=b=c=\frac{1}{3}.$
12.10.2020 03:48
For $ a,b,c>0$ with $a+b+c=3.$ Prove that$$ \sqrt{ab+\frac{3}{8}(a-b)^2}+\sqrt{bc}+\sqrt{ca}\leq 3$$
Attachments:


30.12.2020 16:06
Let $b=\frac{1-a}{2}+\varepsilon$ with $b>c$ WLOG. Then the expression is \[\sqrt{a+\varepsilon^2}+\sqrt{\frac{1-a}{2}+\varepsilon}+\sqrt{\frac{1-a}{2}-\varepsilon}.\]With respect to $\varepsilon$, the derivative of this expression is \[\frac{\varepsilon}{\sqrt{a+\varepsilon^2}}+\frac{1}{2\sqrt{\frac{1-a}{2}+\varepsilon}} - \frac{1}{2\sqrt{\frac{1-a}{2}-\varepsilon}}.\]By computation, we have \[\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}} = \frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}} = \frac{y-x}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}\]for general $x,y$. Thus the derivative of the expression is \[\frac{\varepsilon}{\sqrt{a+\varepsilon^2}}-\frac{\varepsilon}{\sqrt{\frac{(1-a)^2}{4}-\varepsilon^2}\cdot \left(\sqrt{\frac{1-a}{2}-\varepsilon}+\sqrt{\frac{1-a}{2}+\varepsilon}\right)}.\]This derivative is zero exactly when either $\varepsilon=0$ or \[a+\varepsilon^2=\left(\frac{(1-a)^2}{4}-\varepsilon^2\right)\cdot \left(1-a+\sqrt{(1-a)^2-4\varepsilon^2}\right).\]Moreover, the expression is positive when \[a+\varepsilon^2<\left(\frac{(1-a)^2}{4}-\varepsilon^2\right)\cdot \left(1-a+\sqrt{(1-a)^2-4\varepsilon^2}\right)\]and negative when the reverse holds. Thus it suffices to check the result for the extremal values of $\varepsilon$: when it is either $0$ or $\frac{1-a}{2}$. In the case that $\varepsilon=0$, the result follows from Jensen on $\sqrt{x}$. In the case that $\varepsilon=\frac{1-a}{2}$, we wish to show \[\sqrt{3}\ge \sqrt{a+\frac{(1-a)^2}{4}}+\sqrt{1-a} = \sqrt{\frac{a^2-2a+1+4a}{4}}+\sqrt{1-a}=\frac{a+1}{2}+\sqrt{1-a}.\]The derivative of this expression with respect to $a$ is \[\frac{1}{2}-\frac{1}{2\sqrt{1-a}}.\]Since $\sqrt{1-a}\le 1$, this expression is always nonnegative, so it suffices to check $a=0$. For $a=0$, the expression is $\frac{1}{2}+\sqrt{1}=\frac{3}{2}<\sqrt{3}$, so we are done.
31.12.2020 13:00
I saw someone say this is easy with Lagrange multipliers - is the rest of this forum just amazing at differentiating and solving 4 variable equations?! I need LM as a backup plan, so any advice on how people find it so easy is appreciated...
23.08.2021 03:30
For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $$ \sqrt{\sqrt{a}+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}<2$$
28.09.2023 04:13
Squaring both sides of $\sqrt{1-b-c+\tfrac{(b-c)^2}{4}} \le \sqrt{3}-\sqrt{b}-\sqrt{c}$, it suffices to show \[1-b-c+\frac{(b-c)^2}{4} \le b+c+2\sqrt{bc}-2\sqrt{3}(\sqrt{b}+\sqrt{c})+3.\]Let $x=\sqrt{b}+\sqrt{c}$ and $y=\sqrt{bc}$. Notice that $b+c=x^2-2y$ and $(b-c)^2=x^4-4x^2y$, so the above inequality is equivalent to \[-x^4+8x^2-8\sqrt{3}x+8 \ge 4y(2-x^2)\]after some rearranging. Since $b+c \le 1$, Jensen on the concave function $f(x)=\sqrt{x}$ gives $x \le \sqrt{2}$, so $2-x^2 \ge 0$. Since $x^2 \ge 4y$ by AM-GM, it suffices to show \[-x^4+8x^2-8\sqrt{3}x+8 \ge x^2(2-x^2),\]but this rearranges to $2(x-\tfrac{2\sqrt{3}}{3})^2 \ge 0$, as desired.
14.02.2024 05:17
Upon substitution $x=\frac{b+c}2+\sqrt{bc},y=\frac{b+c}2-\sqrt{bc}$ we are given $a+x+y=1$ and we want to show $\sqrt{a+xy}+\sqrt{2x}\le\sqrt3.$ But upon setting $a=1-x-y$ we want to show $\sqrt{(1-x)(1-y)}+\sqrt{2x}\le\sqrt3$ for $x+y\le 1,x\ge y\ge 0.$ But now the left side is clearly $\le \sqrt{1-x}+\sqrt{2x}$ and taking the derivative we see this is maximized at $x=\frac23$ which gives $\sqrt{1-x}+\sqrt{2x}=\sqrt3.$ Equality holds at $x=\frac23,y=0$ which becomes $a=b=c=\frac13.$