Bài 4. Chứng minh các bất đẳng thức:
a) \(5(x-1) < x^5– 1< 5x^4(x-1)\), biết \(x – 1 > 0\)
b) \(x^5+ y^5– x^4y – xy^4≥ 0\), biết \(x + y ≥ 0\)
c) \(\sqrt {4a + 1} + \sqrt {4b + 1} + \sqrt {4c + 1} < 5\) , biết rằng \(a, b, c\) cùng lớn hơn và \(a + b + c = 1\)
a) \(x -1 >5 ⇔ x > 1 ⇒ x^4> x^3> x^2> x > 1\)
\(\Rightarrow {\rm{ }}5{x^4} > {\rm{ }}{x^4} + {\rm{ }}{x^3} + {\rm{ }}{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1{\rm{ }} > {\rm{ }}5\)
\(\Rightarrow {\rm{ }}5{x^4}\left( {x - 1} \right){\rm{ }} > {\rm{ }}\left( {x - 1} \right)({\rm{ }}{x^4} + {\rm{ }}{x^3} + {\rm{ }}{x^2} + {\rm{ }}x{\rm{ }} + {\rm{ }}1){\rm{ }} = {\rm{ }}{x^5} - 1{\rm{ }} > {\rm{ }}5{\rm{ }}\left( {x - 1} \right)\)
b)
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\({{x^5} + {\rm{ }}{y^{5}}-{\rm{ }}{x^4}y{\rm{ }}-{\rm{ }}x{y^4} = {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}y} \right)\left( {{x^4}-{\rm{ }}{x^3}y{\rm{ }} + {\rm{ }}{x^2}{y^2}-{\rm{ }}x{y^3} + {\rm{ }}{y^4}} \right){\rm{ }}-{\rm{ }}xy\left( {{x^{3}} + {\rm{ }}{y^3}} \right)}\)
\({ = {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}y} \right){\rm{ }}\left[ {\left( {{\rm{ }}{x^4}-{\rm{ }}{x^3}y + {\rm{ }}{x^2}{y^2}-{\rm{ }}x{y^3} + {\rm{ }}{y^4}} \right){\rm{ }}-{\rm{ }}xy\left( {{x^2}-{\rm{ }}xy{\rm{ }} + {\rm{ }}{y^2}} \right)} \right]}\)
\({ = {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}y} \right){\rm{ }}\left[ {\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right){\rm{ }} - {\rm{ }}2xy\left( {{x^2} + {y^2}} \right)} \right]}\)
\({ = {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}y} \right){\rm{ }}{{\left( {x{\rm{ }} - {\rm{ }}y} \right)}^2}\left( {{x^2} + {\rm{ }}{y^2}} \right){\rm{ }} \ge {\rm{ }}0}\) do \(x + y ≥ 0; (x - y)^2 ≥ 0, x^2 + y^2≥ 0\)
c)
\(\eqalign{
& {(\sqrt {4a + 1} + \sqrt {4b + 1} + \sqrt {4c + 1} )^2} \cr
& = 4(a + b + c) + 3 + 2\sqrt {4a + 1} \sqrt {4b + 1} + 2\sqrt {4a + 1} \sqrt {4c + 1} + 2\sqrt {4b + 1} \sqrt {4c + 1} \cr
& \le 4(a + b + c) + 3 + (4a + 1) + (4b + 1) + (4a + 1) + (4c + 1) + (4b + 1) + (4c + 1) \cr
& \le 12(a + b + c) + 9 \le 21 \le 25 \cr
& \cr} \)
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