Cómo demostrar la desigualdad $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$

Question

Cómo demostrar la desigualdad $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$ para $a,b,c>0$ y $abc=1$ ?

He intentado probar $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$

Sí, es cierto, $\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$

$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$ $\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$ $\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!) Está mal. Consejos para resolver este problema.

Answer 1

Por lo tanto, queda por demostrar que $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$

Tenemos \begin{align*}LHS-RHS&=(x^3+y^3+z^3)\sum x(x-y)(x-z)+(x^2+y^2+z^2-xy-yz-zx)^3\\ &+7(xy+yz+zx)\sum x^2(x-y)(x-z)\\ &+[x^2+y^2+z^2+3(xy+yz+zx)][x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z)]\\ &+ (x+y+z)[4(x^2+y^2+z^2)+9(xy+yz+zx)]\sum z(x-y)^2\\ &+ (x+y+z)(xy+yz+zx)\sum (x+y+7z)(x-y)^2 \ge 0\end{align*}

Answer 2

Algunas observaciones:

$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$

$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$ $$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)\tag{2}$$

Comparando con la definición original: $$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2},\tag{3}$$

Llegamos a la conclusión de que tenemos que demostrar:

$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \le 3-\frac{3\sqrt{2}}{4}\tag{4}$$

Answer 3

Pista que tenemos con $f(x)=\frac{x}{\sqrt{1+x}}$ obtenemos al diferenciar dos veces $f''(x)=-\frac{1}{4}\frac{x+4}{(x+1)^{5/4}}<0$ para $x>0$ así $f(x)$ es estrictamente cóncavo. tenemos $f(a)+f(b)+f(c)\geq 3f\left(\frac{a+b+c}{3}\right)$ esto es $3\frac{\frac{a+b+c}{3}}{\sqrt{1+\frac{a+b+c}{3}}}$ ajuste $t=\frac{a+b+c}{3}$ tenemos que demostrar que $\frac{t}{\sqrt{1+t}}\geq \frac{\sqrt{2}}{2}$ esto equivale a $2(t-1)(t+1/2)\geq 0$ ahora tenemos $t=\frac{a+b+c}{3}\geq \sqrt[3]{abc}=1$ qed. Así que ahora voy a corregir mi prueba: elevando al cuadrado el lado izquierdo y el lado derecho de la inecuación obtenemos por AM-GM: $\frac{a^2}{1+a}+\frac{b^2}{1+b}+\frac{c^2}{1+c}+\frac{2ab}{\sqrt{1+a}\sqrt{1+b}}+\frac{2ac}{\sqrt{1+a}\sqrt{1+b}}+\frac{2ab}{\sqrt{1+b}\sqrt{1+c}}\geq\frac{a^2}{1+a}+\frac{b^2}{1+b}+\frac{c^2}{1+c}+\frac{4ab}{2+a+b}+\frac{4ac}{2+a+c}+\frac{4bc}{2+b+c}\geq \frac{9}{2}$ tenemos que demostrar que $\frac{4ab}{2+a+b}+\frac{4ac}{2+a+c}+\frac{4bc}{2+b+c}\geq \frac{9}{2}$ con $a=x/y,b=y/z,c=z/x$ obtenemos que esto es equivalente a $2\,{x}^{8}{z}^{2}{y}^{2}+6\,{x}^{8}{z}^{3}y+4\,{x}^{8}{z}^{4}+12\,{x}^ {7}z{y}^{4}+33\,{x}^{7}{z}^{2}{y}^{3}+19\,{x}^{7}{z}^{3}{y}^{2}+6\,{x} ^{7}{z}^{4}y+6\,{x}^{7}{z}^{5}+2\,{x}^{6}{y}^{6}+33\,{x}^{6}{y}^{5}z+ 52\,{x}^{6}{z}^{2}{y}^{4}-10\,{x}^{6}{z}^{3}{y}^{3}-10\,{x}^{6}{z}^{4} {y}^{2}+19\,{x}^{6}{z}^{5}y+2\,{x}^{6}{z}^{6}+6\,{x}^{5}{y}^{7}+19\,{x }^{5}{y}^{6}z-10\,{x}^{5}{y}^{5}{z}^{2}-44\,{x}^{5}{z}^{3}{y}^{4}-70\, {x}^{5}{z}^{4}{y}^{3}-10\,{x}^{5}{z}^{5}{y}^{2}+33\,{z}^{6}{x}^{5}y+4 \,{y}^{8}{x}^{4}+6\,{y}^{7}{x}^{4}z-10\,{y}^{6}{x}^{4}{z}^{2}-70\,{y}^ {5}{x}^{4}{z}^{3}-150\,{x}^{4}{z}^{4}{y}^{4}-44\,{z}^{5}{x}^{4}{y}^{3} +52\,{z}^{6}{x}^{4}{y}^{2}+12\,{z}^{7}{x}^{4}y+6\,{y}^{8}{x}^{3}z+19\, {y}^{7}{x}^{3}{z}^{2}-10\,{y}^{6}{x}^{3}{z}^{3}-44\,{y}^{5}{x}^{3}{z}^ {4}-70\,{z}^{5}{x}^{3}{y}^{4}-10\,{z}^{6}{x}^{3}{y}^{3}+33\,{z}^{7}{x} ^{3}{y}^{2}+2\,{y}^{8}{x}^{2}{z}^{2}+33\,{y}^{7}{x}^{2}{z}^{3}+52\,{y} ^{6}{x}^{2}{z}^{4}-10\,{z}^{5}{y}^{5}{x}^{2}-10\,{z}^{6}{x}^{2}{y}^{4} +19\,{z}^{7}{x}^{2}{y}^{3}+2\,{z}^{8}{x}^{2}{y}^{2}+12\,{y}^{7}x{z}^{4 }+33\,{y}^{6}{z}^{5}x+19\,{y}^{5}{z}^{6}x+6\,{z}^{7}x{y}^{4}+6\,{z}^{8 }x{y}^{3}+2\,{y}^{6}{z}^{6}+6\,{y}^{5}{z}^{7}+4\,{z}^{8}{y}^{4} \geq 0$ ajuste $y=x+u,z=x+u+v$ obtenemos $ \left( 1152\,{u}^{2}+1152\,uv+1152\,{v}^{2} \right) {x}^{10}+ \left( 8064\,{u}^{3}+11952\,{u}^{2}v+10800\,u{v}^{2}+3456\,{v}^{3} \right) {x }^{9}+ \left( 24976\,{u}^{4}+49088\,{u}^{3}v+47712\,{u}^{2}{v}^{2}+ 23600\,u{v}^{3}+4240\,{v}^{4} \right) {x}^{8}+ \left( 45024\,{u}^{5}+ 110584\,{u}^{4}v+122736\,{u}^{3}{v}^{2}+75248\,{u}^{2}{v}^{3}+23512\,u {v}^{4}+2720\,{v}^{5} \right) {x}^{7}+ \left( 52252\,{u}^{6}+154796\,{ u}^{5}v+199453\,{u}^{4}{v}^{2}+143974\,{u}^{3}{v}^{3}+59677\,{u}^{2}{v }^{4}+12612\,u{v}^{5}+956\,{v}^{6} \right) {x}^{6}+ \left( 40732\,{u}^ {7}+142353\,{u}^{6}v+213343\,{u}^{5}{v}^{2}+179155\,{u}^{4}{v}^{3}+ 90285\,{u}^{3}{v}^{4}+26273\,{u}^{2}{v}^{5}+3765\,u{v}^{6}+172\,{v}^{7 } \right) {x}^{5}+ \left( 21556\,{u}^{8}+87644\,{u}^{7}v+152436\,{u}^{ 6}{v}^{2}+148590\,{u}^{5}{v}^{3}+88360\,{u}^{4}{v}^{4}+32032\,{u}^{3}{ v}^{5}+6503\,{u}^{2}{v}^{6}+589\,u{v}^{7}+12\,{v}^{8} \right) {x}^{4}+ \left( 7624\,{u}^{9}+35764\,{u}^{8}v+71988\,{u}^{7}{v}^{2}+81423\,{u} ^{6}{v}^{3}+56605\,{u}^{5}{v}^{4}+24573\,{u}^{4}{v}^{5}+6347\,{u}^{3}{ v}^{6}+842\,{u}^{2}{v}^{7}+38\,u{v}^{8} \right) {x}^{3}+ \left( 1716\, {u}^{10}+9250\,{u}^{9}v+21507\,{u}^{8}{v}^{2}+28214\,{u}^{7}{v}^{3}+ 22861\,{u}^{6}{v}^{4}+11697\,{u}^{5}{v}^{5}+3664\,{u}^{4}{v}^{6}+631\, {u}^{3}{v}^{7}+44\,{u}^{2}{v}^{8} \right) {x}^{2}+ \left( 220\,{u}^{11 }+1363\,{u}^{10}v+3661\,{u}^{9}{v}^{2}+5570\,{u}^{8}{v}^{3}+5252\,{u}^ {7}{v}^{4}+3143\,{u}^{6}{v}^{5}+1165\,{u}^{5}{v}^{6}+244\,{u}^{4}{v}^{ 7}+22\,{u}^{3}{v}^{8} \right) x+12\,{u}^{12}+86\,{u}^{11}v+268\,{u}^{ 10}{v}^{2}+474\,{u}^{9}{v}^{3}+520\,{u}^{8}{v}^{4}+362\,{u}^{7}{v}^{5} +156\,{u}^{6}{v}^{6}+38\,{u}^{5}{v}^{7}+4\,{u}^{4}{v}^{8} \geq 0$ lo cual es cierto.