rightways wrote: Given $a,b,c>0$ such that $$a+b+c+\frac{1}{abc}=\frac{19}{2}$$ What is the greatest value for $a$? By AM-GM $$\frac{19}{2}=a+b+c+\frac{1}{abc}\geq a+\frac{3}{\sqrt[3]a},$$which gives $a\leq8.$

Given $a,b,c,d>0$ such that $a+b+c+d+\frac{1}{abcd}=18.$ Prove that $$a\leq16.$$

sqing wrote: Given $a,b,c,d>0$ such that $a+b+c+d+\frac{1}{abcd}=18.$ Prove that $$a\leq16.$$ Because $$18=\frac{31a}{32}+(\frac{a}{32}+b+c+d+\frac{1}{abcd})\ge \frac{31a}{32}+\frac{5}{2}$$

rightways wrote: sqing wrote: Given $a,b,c,d>0$ such that $a+b+c+d+\frac{1}{abcd}=18.$ Prove that $$a\leq16.$$ Because $$18=\frac{31a}{32}+(\frac{a}{32}+b+c+d+\frac{1}{abcd})\ge \frac{31a}{32}+\frac{5}{2}$$ Very nice. Thanks.

rightways wrote: Given $a,b,c>0$ such that $$a+b+c+\frac{1}{abc}=\frac{19}{2}$$ What is the greatest value for $a$? By AM-GM, $$\frac{19}{2}=\frac{15a}{16}+(\frac{a}{16}+b+c+\frac{1}{abc})\ge\frac{15a}{16}+2$$$$a\leq 8.$$When $a=8, b=c=\frac{1}{2},$ have $a+b+c+\frac{1}{abc}=\frac{19}{2}.$

Given $a,b,c >0$ such that $a+b^2+c^2+\frac{1}{abc}=9.$ Prove that$$a\leq 8.$$Given $a,b,c>0$ such that $a+\frac{4}{3}(b^3+c^3)+\frac{1}{abc}=\frac{53}{6}.$ Prove that$$a\leq8.$$Given $a,b,c,d>0$ such that $a+\frac{4}{3}(b^3+c^3+d^3)+\frac{1}{abcd}=17.$ Prove that$$a\leq16.$$Given $a,b,c,d>0$ such that $a+2(b^4+c^4+d^4)+\frac{1}{abcd}=\frac{135}{8}.$ Prove that$$a\leq16.$$