Mixing
Inequality. $sum_{cyc}(frac{1}{a+b+sqrt{2a+2c}})^3 le frac{8}{9}$
$begingroup$
Problem. When $a, b, c>0, a, b, c in Bbb R, 16(a+b+c)gefrac{1}{a}+frac{1}{b}+frac{1}{c}$, Prove that
$$sum_{cyc}(frac{1}{a+b+sqrt{2a+2c}})^3 le frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) ge -2frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be:
$$sum_{cyc}(frac{1}{x+sqrt{2y}})^3 le frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
inequality substitution a.m.-g.m.-inequality
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
$endgroup$
add a comment |
$begingroup$
Problem. When $a, b, c>0, a, b, c in Bbb R, 16(a+b+c)gefrac{1}{a}+frac{1}{b}+frac{1}{c}$, Prove that
$$sum_{cyc}(frac{1}{a+b+sqrt{2a+2c}})^3 le frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) ge -2frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be:
$$sum_{cyc}(frac{1}{x+sqrt{2y}})^3 le frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
inequality substitution a.m.-g.m.-inequality
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
$endgroup$
$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53
add a comment |
$begingroup$
Problem. When $a, b, c>0, a, b, c in Bbb R, 16(a+b+c)gefrac{1}{a}+frac{1}{b}+frac{1}{c}$, Prove that
$$sum_{cyc}(frac{1}{a+b+sqrt{2a+2c}})^3 le frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) ge -2frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be:
$$sum_{cyc}(frac{1}{x+sqrt{2y}})^3 le frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
inequality substitution a.m.-g.m.-inequality
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
$endgroup$
Problem. When $a, b, c>0, a, b, c in Bbb R, 16(a+b+c)gefrac{1}{a}+frac{1}{b}+frac{1}{c}$, Prove that
$$sum_{cyc}(frac{1}{a+b+sqrt{2a+2c}})^3 le frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that
$$4(x+y+z) ge -2frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be:
$$sum_{cyc}(frac{1}{x+sqrt{2y}})^3 le frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
inequality substitution a.m.-g.m.-inequality
inequality substitution a.m.-g.m.-inequality
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
edited Jan 11 at 10:37
Michael Rozenberg
102k1791195
102k1791195
asked Jan 11 at 8:11
coding1101coding1101
734
asked Jan 11 at 8:11
coding1101coding1101
734
asked Jan 11 at 8:11
coding1101coding1101
734
734
$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53
add a comment |
$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53
$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53
$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a=frac{x}{4},$ $b=frac{y}{4}$ and $c=frac{z}{4}.$
Thus, the condition gives $$x+y+zgeqfrac{1}{x}+frac{1}{y}+frac{1}{z}$$ or
$$1leqfrac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$sum_{cyc}frac{1}{left(x+y+2sqrt{2(x+z)}right)^3}leqfrac{1}{72}.$$
Now, by AM-GM
$$left(x+y+2sqrt{2(x+z)}right)^3geqleft(3sqrt[3]{(x+y)left(sqrt{2(x+z)}right)^2}right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$sum_{cyc}frac{1}{(x+z)(y+z)}leqfrac{3}{4}$$ or
$$8(x+y+z)leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)cdotfrac{xyz(x+y+z)}{xy+xz+yz}leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$sum_{cyc}z(x-y)^2geq0,$$ it's enough to prove that
$$frac{8}{9}(x+y+z)(xy+xz+yz)cdot3(xy+xz+yz)geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z).$$
Can you end it now?
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
active
oldest
votes
$begingroup$
Let $a=frac{x}{4},$ $b=frac{y}{4}$ and $c=frac{z}{4}.$
Thus, the condition gives $$x+y+zgeqfrac{1}{x}+frac{1}{y}+frac{1}{z}$$ or
$$1leqfrac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$sum_{cyc}frac{1}{left(x+y+2sqrt{2(x+z)}right)^3}leqfrac{1}{72}.$$
Now, by AM-GM
$$left(x+y+2sqrt{2(x+z)}right)^3geqleft(3sqrt[3]{(x+y)left(sqrt{2(x+z)}right)^2}right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$sum_{cyc}frac{1}{(x+z)(y+z)}leqfrac{3}{4}$$ or
$$8(x+y+z)leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)cdotfrac{xyz(x+y+z)}{xy+xz+yz}leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$sum_{cyc}z(x-y)^2geq0,$$ it's enough to prove that
$$frac{8}{9}(x+y+z)(xy+xz+yz)cdot3(xy+xz+yz)geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z).$$
Can you end it now?
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
add a comment |
$begingroup$
Let $a=frac{x}{4},$ $b=frac{y}{4}$ and $c=frac{z}{4}.$
Thus, the condition gives $$x+y+zgeqfrac{1}{x}+frac{1}{y}+frac{1}{z}$$ or
$$1leqfrac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$sum_{cyc}frac{1}{left(x+y+2sqrt{2(x+z)}right)^3}leqfrac{1}{72}.$$
Now, by AM-GM
$$left(x+y+2sqrt{2(x+z)}right)^3geqleft(3sqrt[3]{(x+y)left(sqrt{2(x+z)}right)^2}right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$sum_{cyc}frac{1}{(x+z)(y+z)}leqfrac{3}{4}$$ or
$$8(x+y+z)leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)cdotfrac{xyz(x+y+z)}{xy+xz+yz}leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$sum_{cyc}z(x-y)^2geq0,$$ it's enough to prove that
$$frac{8}{9}(x+y+z)(xy+xz+yz)cdot3(xy+xz+yz)geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z).$$
Can you end it now?
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
add a comment |
$begingroup$
Let $a=frac{x}{4},$ $b=frac{y}{4}$ and $c=frac{z}{4}.$
Thus, the condition gives $$x+y+zgeqfrac{1}{x}+frac{1}{y}+frac{1}{z}$$ or
$$1leqfrac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$sum_{cyc}frac{1}{left(x+y+2sqrt{2(x+z)}right)^3}leqfrac{1}{72}.$$
Now, by AM-GM
$$left(x+y+2sqrt{2(x+z)}right)^3geqleft(3sqrt[3]{(x+y)left(sqrt{2(x+z)}right)^2}right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$sum_{cyc}frac{1}{(x+z)(y+z)}leqfrac{3}{4}$$ or
$$8(x+y+z)leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)cdotfrac{xyz(x+y+z)}{xy+xz+yz}leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$sum_{cyc}z(x-y)^2geq0,$$ it's enough to prove that
$$frac{8}{9}(x+y+z)(xy+xz+yz)cdot3(xy+xz+yz)geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z).$$
Can you end it now?
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
$endgroup$
Let $a=frac{x}{4},$ $b=frac{y}{4}$ and $c=frac{z}{4}.$
Thus, the condition gives $$x+y+zgeqfrac{1}{x}+frac{1}{y}+frac{1}{z}$$ or
$$1leqfrac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that
$$sum_{cyc}frac{1}{left(x+y+2sqrt{2(x+z)}right)^3}leqfrac{1}{72}.$$
Now, by AM-GM
$$left(x+y+2sqrt{2(x+z)}right)^3geqleft(3sqrt[3]{(x+y)left(sqrt{2(x+z)}right)^2}right)^3=54(x+y)(x+z).$$
Id est, it's enough to prove that
$$sum_{cyc}frac{1}{(x+z)(y+z)}leqfrac{3}{4}$$ or
$$8(x+y+z)leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that
$$8(x+y+z)cdotfrac{xyz(x+y+z)}{xy+xz+yz}leq3(x+y)(x+z)(y+z),$$ or
$$3(x+y)(x+z)(y+z)(xy+xz+yz)geq8xyz(x+y+z)^2.$$
Now, since $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$sum_{cyc}z(x-y)^2geq0,$$ it's enough to prove that
$$frac{8}{9}(x+y+z)(xy+xz+yz)cdot3(xy+xz+yz)geq8xyz(x+y+z)^2$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z).$$
Can you end it now?
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
answered Jan 11 at 10:35
Michael RozenbergMichael Rozenberg
102k1791195
102k1791195
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
add a comment |
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
Of course. Thanks for answering my question.
$endgroup$
– coding1101
Jan 14 at 8:17
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
$begingroup$
@coding1101 You are welcome!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:41
add a comment |
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$begingroup$
Welcome to Math.SE. We look for more information in posts here than some other sites - in particular, please include the source of problems like this as well as some motivation for them, whenever possible.
$endgroup$
– Carl Mummert
Jan 11 at 13:53