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Inequalities
sqing   5
N an hour ago by sqing
Let $ a,b \geq 0 , a^2+b^2= 3 $ and $ a^4+b^4 = 7.$ Prove that $$\frac{15\sqrt 5-11}{2}\leq  a^5+2b^5\leq \frac{15\sqrt 5+11}{2}$$$$  15\sqrt 5-22 \leq a^5 +5b^5\leq 15\sqrt 5+22 $$
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sqing
Yesterday at 11:56 AM
sqing
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