Solution (not mine). Let $a + b + c = t$, and let $f(x) = \frac{x}{t - x}$. It suffices to prove that $$ \sqrt[7]{f(a)+f(b)}+\sqrt[7]{f(b)+f(c)}+\sqrt[7]{f(c)+f(a)} \geq 3. $$ It is well known that $f(x)$ is convex for $0 < x < t$. Then, applying Jensen's inequality, $$ f(a) + f(b) \geq 2 f\left(\frac{a + b}{2}\right) $$so $$ \sqrt[7]{f(a)+f(b)}+\sqrt[7]{f(b)+f(c)}+\sqrt[7]{f(c)+f(a)} \geq \sqrt[7]{2 f\left(\frac{a + b}{2}\right)} + \sqrt[7]{2 f\left(\frac{b + c}{2}\right)} + \sqrt[7]{2 f\left(\frac{c + a}{2}\right)}. $$ Since $\sqrt[7]{y} < \sqrt[7]{x}$ iff $y < x$, if $g(x) = \sqrt[7]{2f(x)}$, then we know that $g$ is also convex for $0 < x < t$ so using Jensen's inequality on the RHS of the above, we have $$ g\left(\frac{a + b}{2}\right) + g\left(\frac{b + c}{2}\right) + g\left(\frac{c + a}{2}\right) \geq 3g\left(\frac{\frac{a + b}{2} + \frac{b + c}{2} + \frac{c + a}{2}}{3}\right) = 3g\left(\frac{a + b + c}{3}\right) = 3g\left(\frac{t}{3}\right). $$Finally, $$ 3g\left(\frac{t}{3}\right) = 3\sqrt[7]{2f\left(\frac{t}{3}\right)} = 3\sqrt[7]{2 \cdot \frac{t/3}{2t/3}} = 3\sqrt[7]{1} = 1, $$so $$ \sqrt[7]{f(a)+f(b)}+\sqrt[7]{f(b)+f(c)}+\sqrt[7]{f(c)+f(a)} \geq 3, $$as required.

$g$ is not necessarily convex on $(0,t)$, just increasing.

Wrong solution rip: The inequality is homogenous so WLOG assume $a+b+c=3$. Notice that we have equality at $a=b=c=1$. Suppose we set $(a, b, c)=(1+\epsilon, 1-\epsilon, 1)$ , where $0<\epsilon<1$ . We then have: \begin{align*} \sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} &=\sqrt[7]{\frac{1+\epsilon}{2-\epsilon}+\frac{1-\epsilon}{2+\epsilon}}+\sqrt[7]{\frac{1-\epsilon}{2}+\frac 12}+\sqrt[7]{\frac 12 +\frac{1+\epsilon}{2}} \\ &=\sqrt[7]{\frac{4+2\epsilon^2}{4-\epsilon^2}}+\sqrt[7]{1-\frac{\epsilon}{2}}+\sqrt[7]{1+\frac{\epsilon}{2}} \end{align*}By AM-GM: $$\sqrt[7]{\frac{4+2\epsilon^2}{4-\epsilon^2}}+\sqrt[7]{1-\frac{\epsilon}{2}}+\sqrt[7]{1+\frac{\epsilon}{2}}\geq 3\sqrt[21]{1+\frac{\epsilon^2}{2}}>3$$Thus, by perturbing the variables by $\epsilon$, we increase the LHS, so the minimum must be achieved at $a=b=c=1$, and this minimum is 3 as desired.

Let $a, b, c>0 .$ Prove that

$$\sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} $$$$\geq 3\sqrt[21] {(\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq 3.$$h

sqing wrote: Let $a, b, c>0 .$ Prove that

$$\sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} $$$$\geq 3\sqrt[21] {(\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq 3.$$h

The first inequality in the hidden box: what inequality used?

stayhomedomath wrote: sqing wrote:

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The first inequality in the hidden box: what inequality used? Chebychev's inequality

stayhomedomath wrote: sqing wrote: Let $a, b, c>0 .$ Prove that

$$\sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} $$$$\geq 3\sqrt[21] {(\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq 3.$$h

The first inequality in the hidden box: what inequality used? Tchebyshev and AM-GM inequality thank you !

Let $a, b, c>0 .$ Prove that \[\sqrt[3]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[3]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[3]{\frac{c}{a+b}+\frac{a}{b+c}} \geq 3.\]

stayhomedomath wrote: sqing wrote: Let $a, b, c>0 .$ Prove that

$$\sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} $$$$\geq 3\sqrt[21] {(\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq 3.$$h

The first inequality in the hidden box: what inequality used? $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{b+c}{2}\left(\frac{1}{c+a}+ \frac{1}{a+b}\right)\iff b^2+c^2\geq 2bc$$

thepsserby wrote: The inequality is homogenous so WLOG assume $a+b+c=3$. Notice that we have equality at $a=b=c=1$. Suppose we set $(a, b, c)=(1+\epsilon, 1-\epsilon, 1)$ , where $0<\epsilon<1$ . We then have: \begin{align*} \sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} &=\sqrt[7]{\frac{1+\epsilon}{2-\epsilon}+\frac{1-\epsilon}{2+\epsilon}}+\sqrt[7]{\frac{1-\epsilon}{2}+\frac 12}+\sqrt[7]{\frac 12 +\frac{1+\epsilon}{2}} \\ &=\sqrt[7]{\frac{4+2\epsilon^2}{4-\epsilon^2}}+\sqrt[7]{1-\frac{\epsilon}{2}}+\sqrt[7]{1+\frac{\epsilon}{2}} \end{align*}By AM-GM: $$\sqrt[7]{\frac{4+2\epsilon^2}{4-\epsilon^2}}+\sqrt[7]{1-\frac{\epsilon}{2}}+\sqrt[7]{1+\frac{\epsilon}{2}}\geq 3\sqrt[21]{1+\frac{\epsilon^2}{2}}>3$$Thus, by perturbing the variables by $\epsilon$, we increase the LHS, so the minimum must be achieved at $a=b=c=1$, and this minimum is 3 as desired. Doesn’t work.

Notice that $$\frac{a}{b+c}+\frac{b}{c+a}=\frac{a^2+b^2+c(a+b)}{(b+c)(c+a)} \stackrel{QM-AM}{\geq} \frac{(a+b)(a+b+2c)}{2(b+c)(c+a)}$$Then $$\sum_{cyc} \sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}\geq \sum_{cyc} \sqrt[7]{\frac{(a+b)(a+b+2c)}{2(b+c)(c+a)}} \stackrel{AM-GM}{\geq}3 \sqrt[21]{\frac{(2a+b+c)(a+2b+c)(a+b+2c)}{8(a+b)(b+c)(c+a)}}$$Now it suffices to prove that $$(2a+b+c)(a+2b+c)(a+b+2c) \geq 8(a+b)(b+c)(c+a)$$$$2a^3+2b^3+2c^3+16abc+7\sum_{sym}a^2b\geq 16abc+8\sum_{sym}a^2b$$$$2a^3+2b^3+2c^3 \geq a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$$$$\sum_{cyc}(a-b)^2(a+b)\geq 0$$Which is obvious, so we're done. $\square$

What does the “cyc” and the “sym” mean? @snakeaid

"cyc" means cyclic and "sym" means symmetric . For example, $$\sum_{cyc}a^2b=a^2b+b^2c+c^2a$$and $$\sum_{sym}a^2b=a^2b+a^2c+b^2c+b^2a+c^2a+c^2b.$$

thepsserby wrote: The inequality is homogenous so WLOG assume $a+b+c=3$. Notice that we have equality at $a=b=c=1$. Suppose we set $(a, b, c)=(1+\epsilon, 1-\epsilon, 1)$ , where $0<\epsilon<1$. Hello you cannot fix $c$ as $1$

circlethm wrote: Let $a, b, c>0 .$ Prove that \[ \sqrt[7]{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt[7]{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} \geq 3. \] 2021 Azerbaijan