Multiply the inequality by $3$. Then by GM-AM we have $$\sum\sqrt{\frac{9a^2}{7a^2+b^2+c^2}\cdot1}\leq\sum \frac{\frac{9a^2}{7a^2+b^2+c^2}+1}{2}\leq3\iff\sum \frac{3a^2}{7a^2+b^2+c^2} \leq1.$$By subtracting $\frac{1}{2}$ from each fraction and multiplying the inequality by $(-1)$ we have $\sum \frac{\sum a^2}{7a^2+b^2+c^2} \geq 1$, which is true by Titu.

WLOG $a^2+b^2+c^2=3$. Note that $f(x)=\sqrt{\frac{x}{6x+3}}$ is concave. Thus, $$LHS=f(a^2)+f(b^2)+f(c^2)\leq 3f\left(\frac{a^2+b^2+c^2}{3}\right)=1$$by Jensen.

$\frac{a}{\sqrt{7a^2+b^2+c^2}}+\frac{b}{\sqrt{a^2+7b^2+c^2}}+\frac{c}{\sqrt{a^2+b^2+7c^2}} \leq 1$ $\frac{a}{\sqrt{7a^2+b^2+c^2}}=\frac{3a}{\sqrt{(7+1+1)(7a^2+b^2+c^2)}}$ Applying Cauchy-Bunyakovsky inequality, we have $\frac{3a}{\sqrt{(7+1+1)(7a^2+b^2+c^2)}}\leq\frac{3a}{7a+b+c}$ Similarly for other 2 fractions. Summing all 3 inequalities we get $\frac{a}{\sqrt{7a^2+b^2+c^2}}+\frac{b}{\sqrt{a^2+7b^2+c^2}}+\frac{c}{\sqrt{a^2+b^2+7c^2}}\leq\frac{3a}{7a+b+c}+\frac{3b}{7b+c+a}+\frac{3c}{7c+a+b}.$ We need to show that $$\frac{3a}{7a+b+c}+\frac{3b}{7b+c+a}+\frac{3c}{7c+a+b}\leq1$$After multiplying everything with $(7a+b+c)(7b+c+a)(7c+a+b)$ and simplifying we get $a^3+b^3+c^3+3(a^2b+b^2c+c^2a)+3(a^2c+b^2a+c^2b)\geq21abc$ which is evident from AM-GM and also is evident that the AM-GM and Cauchy-Bunyakovsky inequality condition of equality is $a=b=c$

Here is an alternative, essentially for part $\sum_{{\rm cyc}}\frac{3a}{7a+b+c}\leqslant 1$), but I'll post the whole for completeness. You don't need brute force. It suffices to show $$ \sum_{{\rm cyc}}\frac{3a}{\sqrt{9(7a^2+b^2+c^2)}}\leqslant 1. $$Now by Cauchy-Schwarz inequality, we have $$ (7+1+1)(7a^2+b^2+c^2)\geqslant (7a+b+c)^2. $$Hence, $$ \sum_{{\rm cyc}}\frac{3a}{\sqrt{9(7a^2+b^2+c^2)}}\leqslant\frac{3a}{7a+b+c}. $$Thus, it suffices to establish $$ \sum_{{\rm cyc}}\frac{7a}{7a+b+c}\leqslant \frac73. $$Using now the trick $$ \frac{7a}{7a+b+c}=1-\frac{b+c}{7a+b+c}, $$it suffices to establish $$ \sum_{{\rm cyc}}\frac{a+b}{7c+a+b}\geqslant \frac23. $$Now, $$ \frac{a}{7c+a+b}+\frac{b}{7a+b+c}+\frac{c}{7b+c+a}\geqslant \frac{(a+b+c)^2}{a^2+b^2+c^2+8(ab+bc+ca)}, $$by Cauchy-Schwarz. Furthermore, $$ \frac{(a+b+c)^2}{a^2+b^2+c^2+8(ab+bc+ca)}\geqslant \frac13. $$Writing Cauchy-Schwarz once more, we obtain the desired conclusion.

Show that for any positive real numbers $a$, $b$, $c$ the following inequality takes places$$\frac{a}{7a+b+c}+\frac{b}{7b+c+a}+\frac{c}{7c+a+b}\leq\frac{1}{3}$$Solution: WLOG a+b+c=3,

augustin_p wrote: Show that for any positive real numbers $a$, $b$, $c$ the following inequality takes place $\frac{a}{\sqrt{7a^2+b^2+c^2}}+\frac{b}{\sqrt{a^2+7b^2+c^2}}+\frac{c}{\sqrt{a^2+b^2+7c^2}} \leq 1$. See https://artofproblemsolving.com/community/c6h1145950p5403768 Show that for any positive real numbers $a$, $b$, $c$ the following inequality takes place $$\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{a+b+c}\le \sqrt{\frac{a}{7a+b+c}}+\sqrt{\frac{b}{7b+c+a}}+\sqrt{\frac{c}{7c+a+b}}\leq 1 \leq \sqrt{\frac{a}{a+b+7c}}+\sqrt{\frac{b}{b+c+7a}}+\sqrt{\frac{c}{c+a+7b}}$$here here Show that for any positive real numbers $a$, $b$, $c$ the following inequality takes places$$\frac{ab}{a^2+b^2+3c^2}+\frac{bc}{3a^2+b^2+c^2}+\frac{ca}{a^2+3b^2+c^2}\leq\frac{3}{5}$$

Inequality can be homogenised to $a+b+c=1$. So,$f(a,b,c)=LHS$ & $g(a,b,c)=a+b+c-1$ Therefore, $\frac{\partial f}{\partial a}+\lambda \frac{\partial g}{\partial a}=0$ and let $\sqrt{7a^{2}+b^{2}+c^{2}}= x$ $\sqrt{a^{2}+7b^{2}+c^{2}}=y$ $\sqrt{a^{2}+b^{2}+7c^{2}}=z$ Now by evaluating oour equations in terms of $-\lambda$ we get, $x+a(\frac{7}{x}+\frac{1}{y}+\frac{1}{z})=x y+b(\frac{1}{x}+\frac{7}{y}+\frac{1}{z})=z+c(\frac{1}{x}+\frac{1}{y}+\frac{7}{z})=-\lambda$ ,As $a=b=c$ satisfies this equation it will yield the max value of $f$.. By putting $a=b=c$ in $f$ we get $1$.

From Cauchy we get $(\sum_{{\rm cyc}}\frac{a^2}{7a^2+b^2+c^2})(1+1+1) \geq \sum_{{\rm cyc}}\frac{a}{\sqrt{7a^2+b^2+c^2}}$ So we want to show that $\frac{1}{3} \geq \sum_{{\rm cyc}}\frac{a^2}{7a^2+b^2+c^2}$ WLOG $a^2+b^2+c^2=3$ so $\frac{1}{3} \geq \sum_{{\rm cyc}}\frac{a^2}{7a^2+b^2+c^2} \Rightarrow\frac{1}{3} \geq \sum_{{\rm cyc}}\frac{a^2}{6a^2+3} \Rightarrow 1 \geq \sum_{{\rm cyc}}\frac{a^2}{2a^2+1} \Rightarrow 2 \geq \sum_{{\rm cyc}}\frac{2a^2}{2a^2+1} = 3- \sum_{{\rm cyc}}\frac{1}{2a^2+1}$ So it comes down to $\sum_{{\rm cyc}}\frac{1}{2a^2+1} \geq 1$ which is true by AM-HM since $a^2+b^2+c^2=3$

бим бим бам бам $\sum{\frac{a}{7a^2+b^2+c^2}} \le 1$ $\sum{\frac{a}{6a^2+a^2+b^2+c^2}} \le 1$ Бy homogenity $a^2+b^2+c^2 = 3$ $\sum{\frac{a}{6a^2+3}} \le 1$ Бy Cauchy $\sum{\frac {a^2}{2a^2+1}} \le 1$ $\equiv$ $\sum{\frac {1}{2a^2+1}} \geq 1$ Which is true Бy Titu it is done

i guess simple tangent line should work