Prove that $sumlimits_{cyc}sqrt[3]{a^2+4bc}geqsqrt[3]{45(ab+ac+bc)}$
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Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$sqrt[3]{a^2+4bc}+sqrt[3]{b^2+4ac}+sqrt[3]{c^2+4ab}geqsqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$left(sumlimits_{cyc}sqrt[3]{a^2+4bc}right)^3sum_{cyc}(a^2+4bc)^3(ka+b+c)^4geqleft(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4$$
Thus, it remains to prove that
$$left(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4geq45(ab+ac+bc)sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $kgeq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
Thank tou!
inequality contest-math
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
add a comment |
up vote
3
down vote
favorite
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$sqrt[3]{a^2+4bc}+sqrt[3]{b^2+4ac}+sqrt[3]{c^2+4ab}geqsqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$left(sumlimits_{cyc}sqrt[3]{a^2+4bc}right)^3sum_{cyc}(a^2+4bc)^3(ka+b+c)^4geqleft(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4$$
Thus, it remains to prove that
$$left(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4geq45(ab+ac+bc)sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $kgeq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
Thank tou!
inequality contest-math
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$sqrt[3]{a^2+4bc}+sqrt[3]{b^2+4ac}+sqrt[3]{c^2+4ab}geqsqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$left(sumlimits_{cyc}sqrt[3]{a^2+4bc}right)^3sum_{cyc}(a^2+4bc)^3(ka+b+c)^4geqleft(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4$$
Thus, it remains to prove that
$$left(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4geq45(ab+ac+bc)sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $kgeq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
Thank tou!
inequality contest-math
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$sqrt[3]{a^2+4bc}+sqrt[3]{b^2+4ac}+sqrt[3]{c^2+4ab}geqsqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$left(sumlimits_{cyc}sqrt[3]{a^2+4bc}right)^3sum_{cyc}(a^2+4bc)^3(ka+b+c)^4geqleft(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4$$
Thus, it remains to prove that
$$left(sumlimits_{cyc}(a^2+4bc)(ka+b+c)right)^4geq45(ab+ac+bc)sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $kgeq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
Thank tou!
inequality contest-math
inequality contest-math
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
edited Nov 12 at 19:19
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
asked Nov 3 '16 at 8:45
Michael Rozenberg
94.2k1588183
94.2k1588183
add a comment |
add a comment |
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