emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Hence, $\left(\sum_{cyc}\sqrt{a+b}\right)^2=2(a+b+c)+2\sum_{cyc}\sqrt{(a+b)(a+c)}\geq$ $\geq2(a+b+c)+2\sum_{cyc}(a+\sqrt{ab})=12u+2\sum_{cyc}\sqrt{ab}\geq12u+6\sqrt[3]{abc}=12u+6w$. Hence, it remains to prove that $f(w^3)\leq0$, where $f(w^3)=(2+5w^3)^2-6w-12u$. But $f''(w^3)=\left(10(2+5w^3)-\frac{2}{w^2}\right)'=50+\frac{4}{w^5}>0$. Hence, by $uvw$ it remains to prove that $f(w^3)\leq0$ in two following cases: 1. $c=0$, which gives $a+b\geq1$, which is $a^2+b^2+2ab\geq a^2+b^2$, which is obvious. 2. $b=c$, which after homogenization and assuming $b=c=1$ gives $(4(a+2)+6\sqrt[3]{a})(a^2+2)^{2.5}\geq(2(a^2+2)^{1.5}+5a)^2$, which is easy. We see that $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 6abc+2$ is also true.

arqady wrote: $(4(a+2)+6\sqrt[3]{a})(a^2+2)^{2.5}\geq(2(a^2+2)^{1.5}+5a)^2$, which is easy. Could you please explain the rest of it ? By the way, with a little help of a graphing calculator, it can be seen that the "5" on the RHS can be replaced by "7" (Maybe further improvements are also available). But, I have no proper proof for it actually.

emregirgin35 wrote: arqady wrote: $(4(a+2)+6\sqrt[3]{a})(a^2+2)^{2.5}\geq(2(a^2+2)^{1.5}+5a)^2$, which is easy. Could you please explain the rest of it ? Certainly! Let $a=x^3$. Hence, we need to prove that $(4x^3+6x+8)(x^6+2)^{\frac{5}{2}}\geq4(x^6+2)^3+25x^6+20x^3\sqrt{(x^6+2)^3}\Leftrightarrow$ $\Leftrightarrow((4x^3+6x+8)(x^6+2)-20x^3)^2(x^6+2)^3-(4(x^6+2)^3+25x^6)^2\geq0$, which is obvious after expanding.

emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ Without loss of generality, assume that $a \ge b \ge c.$ Using Minkowsky's inequality, we have \[\begin{aligned} \mathrm{LHS}& \ge \sqrt{\left(\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{b}+\sqrt{c}+\sqrt{c}\right)^2} \\ &=\sqrt{4a+2b+4c+4\sqrt{ab}+4\sqrt{bc}}\\ & \ge \sqrt{4a+6b+8c}.\end{aligned}\] Therefore, it suffices to prove that \[4a+6b+8c \ge (5abc+2)^2,\] or \[4a+6b+8c \ge 25a^2b^2c^2+20abc+4.\] Since $a^2+b^2+c^2 \le a^2+a(b+c) \le \left(a+\frac{b+c}{2}\right)^2,$ we have \[4\le 4\left(a+\frac{b+c}{2}\right)=4a+2b+2c.\quad (1)\] On the other hand, it is easy to see that \[25a^2b^2c^2 \le \frac{25}{3\sqrt{3}}abc \le 5abc.\quad (2)\] Using $(1)$ and $(2),$ we only have to prove that \[4b+6c \ge 25abc,\] or \[(a^2+b^2+c^2)(4b+6c) \ge 25abc.\] Now, using the AM-GM and Cauchy-Schwarz inequalities, we have \[\mathrm{LHS} \ge \sqrt{2}a(b+c)(6c+4b) \ge \sqrt{2}\left(\sqrt{6}+2\right)^2abc \ge 25abc.\] So we are done.

we have \[ S=2(x+y+z)(\frac{xy}{x+y}+\frac{xz}{x+z}+\frac{yz}{y+z})\ge 5xyz+2 \] for all $x,y,z\ge 0$ with $xy+yz+zx=1$. Let $a=\sqrt{yz},$ ... So we need to prove \[ LHS\ge S. \] can you help me?

emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ I was surprised that this inequality is problem 6, day 2 for Turkey TST 2014. I think it's not enough interesting and sharp. From the condition, we have $ 0 \leq a,b,c \leq 1.$ Using the Cauchy-Schwarz inequality, we have \[ \begin{aligned} \left(\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c}\right)^2 \ & =2(a+b+c)+2\left[\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\right] \\ & \geq 2(a+b+c)+2\big(a+\sqrt{bc}+b+\sqrt{ca}+c+\sqrt{ab}\big) \\ & =4(a+b+c)+2\big(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\big) \geq 4(a+b+c)+2(ab+bc+ca). \end{aligned}\] Using the AM-GM inequality, we have $ abc \leq \dfrac{(a+b+c)(ab+bc+ca)}{9}.$ It suffices to show that \[ \sqrt{4(a+b+c)+2(ab+bc+ca)} \geq 2+\dfrac{5(a+b+c)(ab+bc+ca)}{9}.\] Setting $ a+b+c=x,\ (1 \leq x \leq \sqrt{3}),$ then $ ab+bc+ca=\dfrac{x^2-1}{2},$ the inequality becomes \[ \sqrt{x^2+4x-1} \geq 2+\dfrac{5x(x^2-1)}{18},\] or \[ \dfrac{(x-1)(x+5)}{\sqrt{x^2+4x-1}+2} \geq \dfrac{5x(x^2-1)}{18},\] or \[ \dfrac{x+5}{\sqrt{x^2+4x-1}+2} \geq \dfrac{5x(x+1)}{18}.\] The last inequality follows from $ \dfrac{x+5}{\sqrt{x^2+4x-1}+2}> \dfrac{4}{3} > \dfrac{5x(x+1)}{18},$ with $ 1 \leq x \leq \sqrt{3}.$

emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ It is not hard. But it's nice! Suppose: $1\ge a \ge b \ge c \ge 0 $ then we have: $\sqrt{b+c} \ge \sqrt{2\sqrt{bc}} =\sqrt[4]{4bc} \ge 5abc$ because it $ \Leftrightarrow a^4b^3c^3 \le \frac{4}{5^4}$ But $a^4b^3c^3 \le \sqrt{\frac{4^4*3^6}{10^{10}}}<\frac{4}{5^4}$ And: $(a+b)(a+c)=a^2+ab+bc+ca \ge a^2+b^2+c^2 =1$ then $\sqrt{a+b}+\sqrt{a+c} \ge 2$

alway_be_with_you wrote: emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ It is not hard. But it's nice! Suppose: $1\ge a \ge b \ge c \ge 0 $ then we have: $\sqrt{b+c} \ge \sqrt{2\sqrt{bc}} =\sqrt[4]{4bc} \ge 5abc$ because it $ \Leftrightarrow a^4b^3c^3 \le \frac{4}{5^4}$ But $a^4b^3c^3 \le \sqrt{\frac{4^4*3^6}{10^{10}}}<\frac{4}{5^4}$ And: $(a+b)(a+c)=a^2+ab+bc+ca \ge a^2+b^2+c^2 =1$ then $\sqrt{a+b}+\sqrt{a+c} \ge 2$ Explain： $a^4b^3c^3 \le \sqrt{\frac{4^4*3^6}{10^{10}}}.$

sqing wrote: alway_be_with_you wrote: emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2$ It is not hard. But it's nice! Suppose: $1\ge a \ge b \ge c \ge 0 $ then we have: $\sqrt{b+c} \ge \sqrt{2\sqrt{bc}} =\sqrt[4]{4bc} \ge 5abc$ because it $ \Leftrightarrow a^4b^3c^3 \le \frac{4}{5^4}$ But $a^4b^3c^3 \le \sqrt{\frac{4^4*3^6}{10^{10}}}<\frac{4}{5^4}$ And: $(a+b)(a+c)=a^2+ab+bc+ca \ge a^2+b^2+c^2 =1$ then $\sqrt{a+b}+\sqrt{a+c} \ge 2$ Explain： $a^4b^3c^3 \le \sqrt{\frac{4^4*3^6}{10^{10}}}.$ certainly! it$ \Leftrightarrow x^4y^3z^3 \le \frac{4^4*3^3*3^3}{10^{10}} $ with $x=a^2,y=b^2,z=c^2$ and $x+y+z=1$

$x^4y^3z^3=4^43^33^3\left(\frac{x}{4}\right)^4\left(\frac{y}{3}\right)^3\left(\frac{z}{3}\right)^3\leq4^43^33^3\left(\frac{4\cdot\frac{x}{4}+3\cdot\frac{y}{3}+3\cdot\frac{z}{3}}{10}\right)^{10}=\frac{4^43^33^3}{10^{10}}$.

arqady wrote: $x^4y^3z^3=4^43^33^3\left(\frac{x}{4}\right)^4\left(\frac{y}{3}\right)^3\left(\frac{z}{3}\right)^3\leq4^43^33^3\left(\frac{4\cdot\frac{x}{4}+3\cdot\frac{y}{3}+3\cdot\frac{z}{3}}{10}\right)^{10}$ $=\frac{4^43^33^3}{10^{10}}$. Thank arqady. $(a^4b^3c^3)^2 =4^43^33^3\left(\frac{a^2}{4}\right)^4\left(\frac{b^2}{3}\right)^3\left(\frac{c^2}{3}\right)^3$ $\leq4^43^33^3\left(\frac{4\cdot\frac{a^2}{4}+3\cdot\frac{b^2}{3}+3\cdot\frac{c^2}{3}}{10}\right)^{10}=\frac{4^4\cdot 3^6}{10^{10}}<(\frac{4}{5^4})^2.$

quykhtn-qa1 wrote: I was surprised that this inequality is problem 6, day 2 for Turkey TST 2014. I think it's not enough interesting and sharp. How about finding the sharpest one? Can anyone find the best constant $\lambda$ such that $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq \lambda abc+2$ for all non-negative real numbers satisfying $a^2+b^2+c^2=1$ ?

crazyfehmy wrote: quykhtn-qa1 wrote: I was surprised that this inequality is problem 6, day 2 for Turkey TST 2014. I think it's not enough interesting and sharp. How about finding the sharpest one? Can anyone find the best constant $\lambda$ such that $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq \lambda abc+2$ for all non-negative real numbers satisfying $a^2+b^2+c^2=1$ ? If you carefully look through the proof by mudok, you find that the candidate you are looking for is $\lambda=9\sqrt{2\sqrt{3}}-6\sqrt{3}$.

http://www.artofproblemsolving.com/community/c6h596634p3540275 Let $ a,b,c$ be real positive numbers satisfying $ a^2+b^2+c^2=1$. Prove that$$\sqrt{a+b} +\sqrt{b+c} +\sqrt{c+a} \geq \sqrt{7(a+b+c)-3}.$$

olimovich wrote: crazyfehmy wrote: quykhtn-qa1 wrote: I was surprised that this inequality is problem 6, day 2 for Turkey TST 2014. I think it's not enough interesting and sharp. How about finding the sharpest one? Can anyone find the best constant $\lambda$ such that $\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq \lambda abc+2$ for all non-negative real numbers satisfying $a^2+b^2+c^2=1$ ? If you carefully look through the proof by mudok, you find that the candidate you are looking for is $\lambda=9\sqrt{2\sqrt{3}}-6\sqrt{3}$. Could someone show a proof of why this $\lambda$ is the required candidate? Thanks!

Let $ a,b,c$ be real positive numbers satisfying $ a^2+b^2+c^2=1$. Prove that$$\sqrt{a+b} +\sqrt{b+c} +\sqrt{c+a} \geq \sqrt{(7+\sqrt{3})(a+b+c)-(3+\sqrt{3})}.$$

Anyone for post 17?

sqing wrote: Let $ a,b,c$ be real positive numbers satisfying $ a^2+b^2+c^2=1$. Prove that$$\sqrt{a+b} +\sqrt{b+c} +\sqrt{c+a} \geq \sqrt{(7+\sqrt{3})(a+b+c)-(3+\sqrt{3})}.$$

emregirgin35 wrote: Prove that for all all non-negative real numbers $a,b,c$ with $a^2+b^2+c^2=1$ \[\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c} \geq 5abc+2.\] here is mine. we need to prove $$\sum_{cyc}{\sqrt{a+b}} \ge 5abc +2$$the equality case is $a=1,b=c=0$. after squaring we need $$2\sum_{cyc}{a}+2\sum_{cyc}{\sqrt{(a+b)(b+c)}} \ge 25(abc)^2+20abc+4$$using CS on $(b+a)(b+c)$ we are left with prowing $$4\sum_{cyc}{a}+2\sum_{cyc}{\sqrt{ab}} \ge 25(abc)^2+20abc+4$$also since $a^2+b^2+c^2=1$ we have $(abc)^2 \le \frac{1}{27}$ so $abc \le \frac{1}{5}$ so we need to prove $$4\sum_{cyc}{a}+2\sum_{cyc}{\sqrt{ab}} \ge 25abc+4$$. now we just prove $$\sum_{cyc}{a} \ge 1+3abc$$after squaring we need $$2\sum_{cyc}{ab} \ge 9(abc)^2+6abc$$or now we have from $AM-GM$ that $$abc \le \frac{1}{\sqrt{27}}$$so $$abc \le \frac 15$$so we just prove $$2\sum_{cyc}{ab} \ge abc(\frac{39}{5})$$so we need to prove $$T=\sum_{cyc}{\frac{1}{a}} \ge (\frac{39}{10})$$by holder we have $$T^2 (a^2+b^2+c^2) \ge 27$$so we just need to prove $$27 \ge( \frac{39}{10})^2$$which is obvious as $39 < 40$. now using the above inequality we are left with $$2\sum_{cyc}{\sqrt{ab}} \ge 13abc$$lets prove $$\sum_{cyc}{\sqrt{ab}} \ge 7abc$$as its stronger. we also have $2ab \le a^2+b^2 \le 1$ so $ab \le \sqrt{\frac{ab}{2}}$ we prove $$\sum_{cyc}{ab} \ge \frac{7}{\sqrt{2}}abc$$also $5 > \frac{7}{\sqrt{2}}$ so we prove $$\sum_{cyc}{\frac{1}{a}} \ge 5$$.which is obvious bc as we said $$ \sum_{cyc}{\frac{1}{a}} \ge \sqrt{27}$$and done. . this inequality sure is something really different.

Could I use "the second order partial derivative of (LHS-RHS) with respect to a is always negative" to show that the function (of a) attains minimum only at a=0 or a=1? Hence (LHS-RHS) attains minimum at (1,0,0) or its permutation, which is 0.