Inequality a+b+c+d + 1/abcd > 18

[nippontakkyu]

If a,b,c,d are positive [tex]a^2+b^2+c^2+d^2=1[/tex], prove that [tex]a+b+c+d +\frac{1}{abcd} \geq 18.[/tex]

[pal702004]

[tex](a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)[/tex]

[tex]ab+ac+ad+bc+bd+cd \ge 6\sqrt[6]{a^3b^3c^3d^3} \Rightarrow (a+b+c+d)^2 \ge 1+12\sqrt{abcd}[/tex]

[tex]a^2+b^2+c^2+d^2 \ge 4\sqrt[4]{a^2b^2c^2d^2} \Rightarrow \sqrt{abcd} \le \frac 1 4[/tex]

[tex]\sqrt{abcd}=x, x\in (0;\frac 1 4][/tex]

[tex]a+b+c+d+\frac{1}{abcd} \ge \sqrt{1+12x}+\frac{1}{x^2}[/tex]

[tex]x \le \frac 1 4 \Rightarrow \sqrt{1+12x}>=1+4x[/tex]

[tex]\operatorname{min} \quad (1+4x+\frac{1}{x^2})=18 \quad at \quad x=\frac 1 4[/tex]