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[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!
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Bài 1:Cho a,b,c là các số dương.( a b + b c + c a ) 2 (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2 ( b a + c b + a c ) 2 1 a \dfrac{1}{a} a 1 1 b \dfrac{1}{b} b 1 1 c \dfrac{1}{c} c 1 a 2 a^2 a 2 b 2 b^2 b 2 c 2 c^2 c 2 d 2 d^2 d 2 3 \sqrt{3} 3
a 2 a^2 a 2 b 2 b^2 b 2 c 2 c^2 c 2 d 2 d^2 d 2 3 \sqrt{3} 3
Có :( a d − b c ) 2 + ( a c + b d ) 2 = ( a 2 + b 2 ) ( c 2 + d 2 ) (ad-bc)^2+(ac+bd)^2=(a^2+b^2)(c^2+d^2) ( a d − b c ) 2 + ( a c + b d ) 2 = ( a 2 + b 2 ) ( c 2 + d 2 ) a d − b c = 1 → 1 + ( a c + b d ) 2 = ( a 2 + b 2 ) ( c 2 + d 2 ) ≥ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) ad-bc=1 \rightarrow 1+(ac+bd)^2=(a^2+b^2)(c^2+d^2)\ge 2\sqrt{(a^2+b^2)(c^2+d^2)} a d − b c = 1 → 1 + ( a c + b d ) 2 = ( a 2 + b 2 ) ( c 2 + d 2 ) ≥ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) → a 2 + b 2 + c 2 + d 2 + a c + b d ≥ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) + a c + b d \rightarrow a^2+b^2+c^2+d^2+ac+bd \ge 2\sqrt{(a^2+b^2)(c^2+d^2)} +ac+bd → a 2 + b 2 + c 2 + d 2 + a c + b d ≥ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) + a c + b d → B D T ↔ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) + a c + b d ≥ 3 \rightarrow BDT \leftrightarrow 2\sqrt{(a^2+b^2)(c^2+d^2)}+ac+bd \ge \sqrt{3} → B D T ↔ 2 ( a 2 + b 2 ) ( c 2 + d 2 ) + a c + b d ≥ 3 ↔ 2 1 + ( a c + b d ) 2 + a c + b d ≥ 3 \leftrightarrow 2\sqrt{1+(ac+bd)^2}+ac+bd \ge \sqrt{3} ↔ 2 1 + ( a c + b d ) 2 + a c + b d ≥ 3 x = a c + b d → p = 2 1 + x 2 + x x=ac+bd \rightarrow p= 2\sqrt{1+x^2}+x x = a c + b d → p = 2 1 + x 2 + x ∣ x ∣ = x 2 < 2 1 + x 2 |x| =\sqrt{x^2}< 2\sqrt{1+x^2} ∣ x ∣ = x 2 < 2 1 + x 2 ∣ x ∣ ≥ − x → p > 0 |x| \ge -x \rightarrow p >0 ∣ x ∣ ≥ − x → p > 0 → p 2 = 4 ( 1 + x 2 ) + 4 x 1 + x 2 + x 2 = ( 1 + x 2 ) + 4 x 1 + x 2 + 4 x 2 + 3 = ( 1 + x 2 + 2 x ) 2 + 3 ≥ 3 \rightarrow p^2 =4(1+x^2)+4x\sqrt{1+x^2}+x^2=(1+x^2)+4x\sqrt{1+x^2}+4x^2+3 = (\sqrt{1+x^2}+2x)^2+3 \ge 3 → p 2 = 4 ( 1 + x 2 ) + 4 x 1 + x 2 + x 2 = ( 1 + x 2 ) + 4 x 1 + x 2 + 4 x 2 + 3 = ( 1 + x 2 + 2 x ) 2 + 3 ≥ 3 → p ≥ 3 → S ≥ 3 \rightarrow p \ge \sqrt{3}\rightarrow S \ge \sqrt{3} → p ≥ 3 → S ≥ 3
Bài 1:a b + b c + c a ≥ 3 a b . b c . c a 3 = 3 \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \ge 3\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}} =3 b a + c b + a c ≥ 3 3 b a . c b . a c = 3 b a + c b + a c ≥ 3 b a . c b . a c 3 = 3 \dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c} \ge 3\sqrt[3]{\dfrac{b}{a}.\dfrac{c}{b}.\dfrac{a}{c}} =3 a b + b c + c a ≥ 3 3 a b . b c . c a = 3 ( 1 + 1 + 1 ) ( a 2 b 2 + b 2 c 2 + c 2 a 2 ) ≥ ( a b + b c + c a ) 2 (1+1+1)(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}) \ge (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2 ( 1 + 1 + 1 ) ( b 2 a 2 + c 2 b 2 + a 2 c 2 ) ≥ ( b a + c b + a c ) 2 ⟺ a 2 b 2 + b 2 c 2 + c 2 a 2 ≥ ( a b + b c + c a ) 2 3 \iff \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \ge \dfrac{(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2}{3} ⟺ b 2 a 2 + c 2 b 2 + a 2 c 2 ≥ 3 ( b a + c b + a c ) 2 ⟺ a 2 b 2 + b 2 c 2 + c 2 a 2 ≥ ( a b + b c + c a ) 2 a b + b c + c a \iff \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \ge \dfrac{(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2}{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}} ⟺ b 2 a 2 + c 2 b 2 + a 2 c 2 ≥ b a + c b + a c ( b a + c b + a c ) 2 ⟺ a 2 b 2 + b 2 c 2 + c 2 a 2 ≥ a b + b c + c a \iff \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} ⟺ b 2 a 2 + c 2 b 2 + a 2 c 2 ≥ b a + c b + a c a 2 b 2 + b 2 c 2 + c 2 a 2 + b a + c b + a c ≥ a b + b c + c a + 3 \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3 b 2 a 2 + c 2 b 2 + a 2 c 2 + a b + b c + c a ≥ b a + c b + a c + 3 ⟺ a 2 b 2 + b 2 c 2 + c 2 a 2 + 2 b a + 2 c b + 2 a c ≥ a b + b c + c a + 3 + b a + c b + a c \iff \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{2b}{a}+\dfrac{2c}{b}+\dfrac{2a}{c} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3+\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c} ⟺ b 2 a 2 + c 2 b 2 + a 2 c 2 + a 2 b + b 2 c + c 2 a ≥ b a + c b + a c + 3 + a b + b c + c a ⟺ ( a b + b c + c a ) 2 ≥ ( a + b + c ) ( 1 a + b c + c a ) \iff (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2 \ge (a+b+c)(\dfrac{1}{a}+\dfrac{b}{c}+\dfrac{c}{a}) ⟺ ( b a + c b + a c ) 2 ≥ ( a + b + c ) ( a 1 + c b + a c )
Bài 1:Cho a,b,c là các số dương.( a b + b c + c a ) 2 (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2 ( b a + c b + a c ) 2 1 a \dfrac{1}{a} a 1 1 b \dfrac{1}{b} b 1 1 c \dfrac{1}{c} c 1
Cách khác :( a b + b c + c a ) 2 ≥ ( a + b + c ) ( 1 a + 1 b + 1 c ) (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2\ge(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}) ( b a + c b + a c ) 2 ≥ ( a + b + c ) ( a 1 + b 1 + c 1 ) ↔ a 2 b 2 + b 2 c 2 + c 2 a 2 + 2 ( a c + b a + c b ) ≥ 3 + a b + b c + c a + a c + b a + c b \leftrightarrow \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+2(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b})\ge 3+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b} ↔ b 2 a 2 + c 2 b 2 + a 2 c 2 + 2 ( c a + a b + b c ) ≥ 3 + b a + c b + a c + c a + a b + b c ↔ a 2 b 2 + b 2 c 2 + c 2 a 2 + a c + b a + c b ≥ 3 + a b + b c + c a \leftrightarrow \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge 3+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} ↔ b 2 a 2 + c 2 b 2 + a 2 c 2 + c a + a b + b c ≥ 3 + b a + c b + a c ( a b ; b c ; c a ) = ( x ; y ; z ) \bigl(\begin{smallmatrix} & \dfrac{a}{b} ; \dfrac{b}{c} ;\dfrac{c}{a}& \\ \end{smallmatrix}\bigr) =(x;y;z) ( b a ; c b ; a c ) = ( x ; y ; z ) → B D T ↔ x 2 + y 2 + z 2 + 1 x + 1 y + 1 z ≥ 3 + x + y + z \rightarrow BDT \leftrightarrow x^2+y^2+z^2+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 3+x+y+z → B D T ↔ x 2 + y 2 + z 2 + x 1 + y 1 + z 1 ≥ 3 + x + y + z x 2 + y 2 + z 2 ≥ ( x + y + z ) 2 3 ; 1 x + 1 y + 1 z ≥ 9 x + y + z x^2+y^2+z^2 \ge \dfrac{(x+y+z)^2}{3} ;\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \dfrac{9}{x+y+z} x 2 + y 2 + z 2 ≥ 3 ( x + y + z ) 2 ; x 1 + y 1 + z 1 ≥ x + y + z 9 x + y + z = t ( t > 0 ) → x+y+z=t (t>0)\rightarrow x + y + z = t ( t > 0 ) → 1 3 t 2 + 9 t ≥ 3 + t \dfrac{1}{3}t^2+\dfrac{9}{t}\ge 3+t 3 1 t 2 + t 9 ≥ 3 + t ↔ t 3 − 3 t 2 − 9 t + 27 ≥ 0 \leftrightarrow t^3-3t^2-9t+27\ge 0 ↔ t 3 − 3 t 2 − 9 t + 2 7 ≥ 0 ↔ ( t + 3 ) ( t − 3 ) 2 ≥ 0 \leftrightarrow (t+3)(t-3)^2 \ge 0 ↔ ( t + 3 ) ( t − 3 ) 2 ≥ 0 t > 0 ) t>0) t > 0 )
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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