parmenides51 wrote: Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$For what values of $x$ does the equality hold? Isolated fudging: It suffices to prove that, for all $x > 0$, $$\sqrt{\frac{1}{3x + 1}} \ge \frac{1}{x^{3/4} + 1}.$$which is true since $(x^{3/4} + 1)^2 \ge 3x + 1$ (AM-GM after expanding).

parmenides51 wrote: Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$For what values of $x$ does the equality hold? https://artofproblemsolving.com/community/c275h100419p567300 https://artofproblemsolving.com/community/c6h526120p2982013

sqing wrote: parmenides51 wrote: Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$For what values of $x$ does the equality hold? https://artofproblemsolving.com/community/c275h100419p567300 https://artofproblemsolving.com/community/c6h526120p2982013 You posted this many years ago? $$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$

dragonheart6 wrote: sqing wrote: parmenides51 wrote: Prove that for all real $x > 0$ holds the inequality $$\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{x}{x+3}}\ge 1.$$For what values of $x$ does the equality hold? https://artofproblemsolving.com/community/c275h100419p567300 https://artofproblemsolving.com/community/c6h526120p2982013 You posted this many years ago? $$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$ https://artofproblemsolving.com/community/c6h487873p3074208

sqing wrote: dragonheart6 wrote: You posted this many years ago? $$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$ https://artofproblemsolving.com/community/c6h487873p3074208 It seems there are no nice proofs.

dragonheart6 wrote: sqing wrote: dragonheart6 wrote: You posted this many years ago? $$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$ https://artofproblemsolving.com/community/c6h487873p3074208 It seems there are no nice proofs. Just found that the isolated fudging also works. It suffices to prove that, for all $x > 0$, $$\sqrt[3]{\frac{1}{1+7x}} - \sqrt{\frac{1}{1+3x}} \ge \frac{2x^2 - 2x}{x^3 + 23x^2 + 23x + 1}.$$

Squaring and simplifying, the inequality becomes $2 \sqrt\frac{x}{(3x+1)(x+3)}\geq\frac{8x}{(3x+1)(x+3)}$. Squaring again and simplifying, the inequality reduces to $3x^2-6x+3\geq0$ which is trivial.

Prove that for all real $x > 0$ holds the inequality $$\frac{1}{3x+1}+\sqrt{\frac{x}{x+3}}\ge \frac{3}{4}$$$$\frac{1}{3x+1}+\sqrt[3]{\frac{x}{x+3}}\ge \frac{7}{8}$$

sqing wrote: Prove that for all real $x > 0$ holds the inequality $$\frac{1}{3x+1}+\sqrt{\frac{x}{x+3}}\ge \frac{3}{4}$$

Attachments:

Let $x > 0 .$ Prove that $$\frac{x}{x+1}+\sqrt[3]{\frac{2}{x+1}}\le 1+\frac{2\sqrt 6}{9}$$$$\frac{x}{x+1}+\sqrt[3]{\frac{3}{x+1}}\le \frac{5}{3}$$$$\sqrt[3]{\frac{x}{x+1}}+\frac{3}{x+1} \le \frac{29}{9}$$

dragonheart6 wrote: sqing wrote: dragonheart6 wrote: You posted this many years ago? $$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$ https://artofproblemsolving.com/community/c6h487873p3074208 It seems there are no nice proofs. Let $a=1+x,b=1-x$. If $0\le x<0.3$ , then $\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\ge\sqrt[3]{\frac{(\sqrt{a}+\sqrt{b})^4}{a^2+14ab+b^2}}\ge\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}} $; If $0.3\le x<1$ , then $\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+3\sqrt[3]{\frac{a}{a+7b}\cdot \frac{b}{b+7a}}}$ ${\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+\frac{3}{4}\cdot \frac{192ab}{7a^2+178ab+7b^2}}}$ $\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+3\sqrt[3]{\frac{ab(49a^4+2268a^3b+7654a^2b^2+2268ab^3+49b^4)}{(a+7b)^2(b+7a)^2(7a^2+178ab+7b^2)}}}$ $\ge\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}} $. (Yanwenlan) https://artofproblemsolving.com/community/c6h487873p3112235 https://math.stackexchange.com/questions/2170649/prove-that-sqrt3-fracaa7b-sqrt3-fracbb7a-geq-sqrt-fraca?noredirect=1

sqing wrote: Let $a=1+x,b=1-x$. If $0\le x<0.3$ , then $\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\ge\sqrt[3]{\frac{(\sqrt{a}+\sqrt{b})^4}{a^2+14ab+b^2}}\ge\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}} $; If $0.3\le x<1$ , then $\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+3\sqrt[3]{\frac{a}{a+7b}\cdot \frac{b}{b+7a}}}$ ${\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+\frac{3}{4}\cdot \frac{192ab}{7a^2+178ab+7b^2}}}$ $\ge\sqrt[3]{\frac{a}{a+7b}+\frac{b}{b+7a}+3\sqrt[3]{\frac{ab(49a^4+2268a^3b+7654a^2b^2+2268ab^3+49b^4)}{(a+7b)^2(b+7a)^2(7a^2+178ab+7b^2)}}}$ $\ge\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}} $. (Yanwenlan) https://artofproblemsolving.com/community/c6h487873p3112235 https://math.stackexchange.com/questions/2170649/prove-that-sqrt3-fracaa7b-sqrt3-fracbb7a-geq-sqrt-fraca?noredirect=1 It seems this step is not true (check $x = 1/5$, so $a = 6/5, b = 4/5$): $$\sqrt[3]{\frac{(\sqrt{a}+\sqrt{b})^4}{a^2+14ab+b^2}}\ge\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}$$

Let $x\ge 0 .$ Prove that $$1\le \sqrt{\frac{1}{3x+1}}+\sqrt{\frac{ x}{x+2}} \le \sqrt{\frac{6}{5}}$$$$1\le\sqrt{\frac{1}{3x+1}}+\sqrt{\frac{ x}{x+1}} \le \sqrt{2(\sqrt{3}-1)}$$$$1\le \frac{1}{x+1} + \sqrt[3]{\frac{x}{x+1}} \le 1+\frac{2}{3\sqrt 3}$$$$ \frac{7}{8}\le \frac{x}{x+3} + \sqrt[3]{\frac{1}{3x+1}} \le \frac{\sqrt{33+192\sqrt 3}-1}{16}$$