Jafet98 wrote: Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}\leq\frac{3\sqrt{2}}{4}$$ $$\sum a\sqrt{a^2+6bc}=\frac{1}{\sqrt{2}}\sum\sqrt{2a^2}\cdot \sqrt{a^2+6bc}\leq \frac{1}{2\sqrt{2}}(3\sum a^2+6\sum bc)=\frac{3\sqrt{2}}{4}.$$

Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<\frac{3\sqrt{2}}{4}$$

sqing wrote: Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<\frac{3\sqrt{2}}{4}$$ $\sum_{cyc} a\sqrt{a^2 + 6bc} \le \sqrt{\left(\sum{a^3}\right) + 18abc}\le\sqrt{(a+b+c)^3 - 6abc}\le 1 < \frac{3\sqrt 2}{4}$

Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<1.$$

Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Prove that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}\leq\frac{\sqrt{7}}{3}.$$

sqing wrote: Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Prove that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}\leq\frac{\sqrt{7}}{3}.$$ try $a=1,b=c=0$.

sqing wrote:

Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<1.$$ I think the following inequality is better Let $a,\ b,\ c \ge 0$ such that $a+b+c=1$. Show that $$a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab} \le 1.$$

Let $x=ab+bc+ca$ and $M=\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}$ Cauchy gets $(a^2+b^2+c^2)[(a^2+6bc)+(b^2+6ca)+(c^2+6ab)]\geq M^2$, which is equivalent to $(1-2x)(1+4x)\geq M^2$. It is easy to see that $(1-2x)(1+4x)\leq\frac{9}{8}$ because it is the same as $(8x-1)^2\geq 0$, so $M^2\leq\frac{9}{8}$, which provides the desired result.

abdelkrim wrote: sqing wrote: Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<1.$$ I think the following inequality is better Let $a,\ b,\ c \ge 0$ such that $a+b+c=1$. Show that $$a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab} \le 1.$$ Let $a ,b,c$ be the sides of $ \triangle ABC $. Prove that $$a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}\le(a+b+c)^2.$$Let $a,\ b,\ c> 0 .$ Show that $$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1,$$$$\frac{b+c}{\sqrt{a^2 + 8bc}} + \frac{c+a}{\sqrt{b^2 + 8ca}} + \frac{a+b}{\sqrt{c^2 + 8ab}} \geq 2, $$$$ \sqrt{a^2 + 8bc} +\sqrt{b^2 + 8ca} +\sqrt{c^2 + 8ab} \leq 3(a+b+c).$$Interesting .

sqing wrote: abdelkrim wrote: sqing wrote: Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$. Show that $$a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}<1.$$ I think the following inequality is better Let $a,\ b,\ c \ge 0$ such that $a+b+c=1$. Show that $$a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab} \le 1.$$ Let $a ,b,c$ be the sides of $ \triangle ABC $. Prove that $$a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}\le(a+b+c)^2.$$Let $a,\ b,\ c> 0 .$ Show that $$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1,$$$$\frac{b+c}{\sqrt{a^2 + 8bc}} + \frac{c+a}{\sqrt{b^2 + 8ca}} + \frac{a+b}{\sqrt{c^2 + 8ab}} \geq 2, $$$$ \sqrt{a^2 + 8bc} +\sqrt{b^2 + 8ca} +\sqrt{c^2 + 8ab} \leq 3(a+b+c).$$Interesting . By C-S,have \[\sum{a^2(a^2+8bc)}(1+1+1)\geq (\sum{a\sqrt{a^2+8bc}})^2\]Need to prove \[16\,{s}^{4}\geq3\,{a}^{2} \left( {a}^{2}+8\,bc \right) +3\,{b}^{2} \left( {b}^{2}+8\,ac \right) +3\,{c}^{2} \left( {c}^{2}+8\,ab \right) \]<=> \[bds=10\,{s}^{4}-36\,r \left( 4\,R-r \right) {s}^{2}-6\,{r}^{2} \left( 4\,R +r \right) ^{2}\geq 0\]But $bds=160\, \left( R-2\,r \right) ^{2}{r}^{2}+10\, \left( {s}^{2}-16\,Rr+5\, {r}^{2} \right) ^{2}+32\, \left( R-2\,r \right) \left( {s}^{2}-16\,Rr +5\,{r}^{2} \right) r+288\, \left( R-2\,r \right) {r}^{3}+144\, \left( {s}^{2}-16\,Rr+5\,{r}^{2} \right) Rr\geq 0 $

Hard is this: In triangle,prove that \[\sqrt {{a}^{2}+8\,bc}+\sqrt {{b}^{2}+8\,ac}+\sqrt {{c}^{2}+8\,ab}\geq 2\, \sqrt {3} \left( \sqrt {s \left( s-a \right) }+\sqrt {s \left( s-b \right) }+\sqrt {s \left( s-c \right) } \right) \]

Any solution?

xzlbq wrote: By C-S,have \[\sum{a^2(a^2+8bc)}(1+1+1)\geq (\sum{a\sqrt{a^2+8bc}})^2\]Need to prove \[(a+b+c)^4 \geq 3\sum\limits_{cyc} a^2 (a^2+8bc)\]But \[\text{LHS}-\text{RHS}=\sum\limits_{cyc} {\frac {3\, \left( a-b \right) ^{2} \left( -3\,c+a+b \right) ^{2}}{20} }+\frac{1}{20} \sum\limits_{cyc} \, \left( a+b-c\right) \left( 11\,a+11\,b+17\,c \right) \left( a+b-2\,c \right) ^{2} \geq 0\]

Generalization 1 Let $a,b,c$ be posituve reels such that $a+b+c=1$. Then prove that $$a\sqrt{pa^2+kbc}+b\sqrt{pb^2+kac}+c\sqrt{pc^2+kab}\leq\dfrac{k\sqrt{\dfrac{k}{2}-p}}{2\left(k-2p\right)}$$

Also, we substitute a=tan(A/2)*tan(B/2), b=tan(B/2)*tan(C/2),c=tan(C/2)*tan(A/2), for the above nice inequalities...

This follows easily from \begin{align*} \sum_{cyc}a\sqrt{a^2+6bc}&\leq \sqrt{\left(\sum_{cyc}a\right)\left(\sum_{cyc}a(a^2+6bc)\right)}\\ &\leq\sqrt{\sum_{cyc}a^3+8abc}\\ &=\sqrt{a^3+b^3+c^3+24abc}\\ &\leq\sqrt{(a+b+c)^3}\\ &\leq 1\\ &\le\frac{3\sqrt{2}}{4} \end{align*}