Mixing
minimum value of $(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2$
$begingroup$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2geq 3cdot 2sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2geq 18sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
inequality
asked Jan 17 at 14:55
DXTDXT
5,7752731
$endgroup$
add a comment |
$begingroup$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2geq 3cdot 2sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2geq 18sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
inequality
asked Jan 17 at 14:55
DXTDXT
5,7752731
$endgroup$
add a comment |
$begingroup$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2geq 3cdot 2sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2geq 18sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
inequality
asked Jan 17 at 14:55
DXTDXT
5,7752731
$endgroup$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2geq 3cdot 2sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)cdot (a^{-1}+b^{-1}+c^{-1})^2geq 18sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
inequality
inequality
asked Jan 17 at 14:55
DXTDXT
5,7752731
asked Jan 17 at 14:55
DXTDXT
5,7752731
asked Jan 17 at 14:55
DXTDXT
5,7752731
asked Jan 17 at 14:55
DXTDXT
5,7752731
asked Jan 17 at 14:55
DXTDXT
5,7752731
5,7752731
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Apply $AM ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $le$ quadratic mean”:
$$
frac{4}{frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}}
le sqrt{frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
$endgroup$
add a comment |
$begingroup$
Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Apply $AM ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $le$ quadratic mean”:
$$
frac{4}{frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}}
le sqrt{frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
$endgroup$
add a comment |
$begingroup$
Hint: Apply $AM ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $le$ quadratic mean”:
$$
frac{4}{frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}}
le sqrt{frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
$endgroup$
add a comment |
$begingroup$
Hint: Apply $AM ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $le$ quadratic mean”:
$$
frac{4}{frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}}
le sqrt{frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
$endgroup$
Hint: Apply $AM ge GM$ not to $8a^2 + b^2 + c^2$, but to
$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$
and $HM le GM$ not to $a^{-1}+b^{-1}+c^{-1}$, but to
$$
frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}
$$
The “partitions” are chosen in such a way that equality can hold simultaneously in both estimates, in this case when $2a=b=c$.
One can also use the relationships between generalized means, here “harmonic mean $le$ quadratic mean”:
$$
frac{4}{frac{1}{2a} + frac{1}{2a} + frac{1}{b} + frac{1}{c}}
le sqrt{frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
edited Jan 17 at 15:50
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
answered Jan 17 at 15:09
Martin RMartin R
29.3k33558
29.3k33558
add a comment |
add a comment |
$begingroup$
Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
add a comment |
$begingroup$
Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
add a comment |
$begingroup$
Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
$endgroup$
Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
answered Jan 17 at 15:57
MacavityMacavity
35.5k52554
35.5k52554
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
add a comment |
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Yes! – Why did I not think of that?
$endgroup$
– Martin R
Jan 17 at 16:00
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
$begingroup$
Essentially convexity - whether AM-GM-HM or Hölder or Jensen... so not very different after all.
$endgroup$
– Macavity
Jan 17 at 16:02
1
1
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
$begingroup$
Yes, but this approach works with arbitrary coefficients and does not require to find a suitable partition of the sums.
$endgroup$
– Martin R
Jan 17 at 16:07
add a comment |
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