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Học sinh chăm học
Thành viên
áp dung bdt
4 ( a 3 + b 3 ) 3 ≥ a + b \sqrt[3]{4(a^{3}+b^{3})}\geq a+b 3 4 ( a 3 + b 3 ) ≥ a + b
thât vay
⇔ 4 ( a 3 + b 3 ) ≥ a 3 + b 3 + 3 a b ( a + b ) ⇔ a 3 + b 3 ≥ a b ( a + b ) ⇔ ( a + b ) ( a 2 − a b + b 2 − a b ) ≥ 0 ⇔ ( a + b ) ( a − b ) 2 ≥ 0 ( l u o ^ n d u n g ) \Leftrightarrow 4(a^{3}+b^{3})\geq a^{3}+b^{3}+3ab(a+b)\Leftrightarrow a^{3}+b^{3}\geq ab(a+b)\Leftrightarrow (a+b)(a^{2}-ab+b^{2}-ab)\geq 0\Leftrightarrow (a+b)(a-b)^{2}\geq 0(luôndung) ⇔ 4 ( a 3 + b 3 ) ≥ a 3 + b 3 + 3 a b ( a + b ) ⇔ a 3 + b 3 ≥ a b ( a + b ) ⇔ ( a + b ) ( a 2 − a b + b 2 − a b ) ≥ 0 ⇔ ( a + b ) ( a − b ) 2 ≥ 0 ( l u o ^ n d u n g )
vt
≥ a + b + b + c + a + c a + b + c = 2 \geq \frac{a+b+b+c+a+c}{a+b+c}=2 ≥ a + b + c a + b + b + c + a + c = 2
dau = xay ra khi a=b=c
tìm min P=
4 ( a 3 + b 3 ) 3 + 4 ( b 3 + c 3 ) 3 + 4 ( c 3 + a 3 ) 3 a + b + c \frac{\sqrt[3]{4(a^3+b^3)}+\sqrt[3]{4(b^3+c^3)}+\sqrt[3]{4(c^3+a^3)}}{a+b+c} a + b + c 3 4 ( a 3 + b 3 ) + 3 4 ( b 3 + c 3 ) + 3 4 ( c 3 + a 3 ) với a,b,c>0