CMR: \(\forall a,b,c\in R\) ta có \(\left|a-b\right|+\left|b-c\right|\ge\left|a-c\right|\)
Chứng minh bất đẳng thức:
\(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\forall a,b,c\in R\)
Bất đẳng thức cần chứng minh tương đương:
\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)
\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))
\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).
Vậy ta có đpcm.
\(a^8+b^8-a^6b^2-a^2b^6=\left(a^8-a^6b^2\right)+\left(b^8-a^2b^6\right)=a^6\left(a^2-b^2\right)+b^6\left(b^2-a^2\right)=\left(a^6-b^6\right)\left(a^2-b^2\right)\) nên suy ra được như vậy Quỳnh Anh
CMR: \(\left(1+\frac{a+b+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\ge8\sqrt{abc}\) \(\forall a,b,c\ge0\)
Sử dụng BĐT: \(\left(x+y+z\right)^3\ge27xyz\Rightarrow\left(\frac{x+y+z}{3}\right)^3\ge xyz\)
\(\Rightarrow\left(\frac{1+a+1+b+1+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge3\sqrt[3]{\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cộng vế với vế:
\(1\ge\frac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
Dấu "=" 3 BĐT trên xảy ra khi \(a=b=c\)
Lại có:
\(1+\sqrt[3]{abc}\ge2\sqrt{\sqrt[3]{abc}}\Rightarrow\left(1+\sqrt[3]{abc}\right)^3\ge\left(2\sqrt{\sqrt[3]{abc}}\right)^3=8\sqrt{abc}\)Dấu "=" xảy ra khi \(a=b=c=1\)
Hãy chứng min rằng :
1) \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2},\forall a,b,c,d\in R\)
2) \(\sqrt{4\cos^2x.\cos^2y+\sin^2\left(x-y\right)}+\sqrt{4\sin^2x.\sin^2y+\sin^2\left(x-y\right)}\ge2,\forall x,y\in R\)
1) Bất đẳng thức cần chứng minh
\(\Leftrightarrow\) a2 + b2 + c2 + d2 + \(2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow\) \(ac+bd\le\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\left(1\right)\)
Nếu : ac + bd < 0 : BĐT luôn đúng
Nếu : ac + bd \(\ge\) 0 : Thì (1) tương đương
( ac + bd )2 \(\le\) ( a2 + b2 )( c2 + d2 )
\(\Leftrightarrow\) \(\left(ac\right)^2+\left(bd\right)^2+2abcd\le\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\)
\(\Leftrightarrow\) \(\left(ad\right)^2+\left(bc\right)^2-2abcd\ge0\)
\(\Leftrightarrow\) \(\left(ad-bc\right)^2\ge0\) , luôn đúng , vậy bài toán được chứng minh
2) Chọn :\(\left\{{}\begin{matrix}a=2\cos x.\cos y\\c=2\sin x.\sin y\\b=d=\sin\left(x-y\right)\end{matrix}\right.\)
Từ câu 1) ta có :
\(\sqrt{4\cos^2x.\cos^2y+\sin^2\left(x-y\right)}+\sqrt{4\sin^2x.\sin^2y+\sin^2\left(x-y\right)}\)
\(\ge\sqrt{\left(2\cos x.\cos y+2\sin x.\sin y\right)^2+\left(2\sin\left(x-y\right)\right)^2}\)
\(\ge\sqrt{4\cos^2\left(x-y\right)+4\sin^2\left(x-y\right)}=2\)
Cho a;b;c\(\in\)R .CMR \(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
với ∀a,b,c thuộc R, CMR:
\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\ge2+\frac{2\left(a+b+c\right)}{\sqrt[3]{abc}}\)
Cho a,b,c là các số thực bất kì. CMR:
\(\left|a\right|+\left|b\right|+\left|c\right|+\left|a+b+c\right|\ge\left|a+b\right|+\left|b+c\right|+\left|c+a\right|\)
Một bạn nào đó (không biết rõ tên) đã giải như sau:
Ta có: \(\left|a\right|+\left|b\right|+\left|c\right|+\left|a+b+c\right|\ge\left|a+b\right|+\left|a+c\right|+\left|b\right|+\left|c\right|\)
\(\ge\left|a+b\right|+\left|b+c\right|+\left|c+a\right|\)
Em hãy cho nhận xét về lời giải trên! Nếu sai hãy sửa lại cho đúng!
CMR:\(\forall a,b,c\)ta luôn có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Áp dụng BĐT Cô si, ta có :
\(a^4+b^4\ge2a^2b^2\)
\(b^4+c^4\ge2b^2c^2\)
\(c^4+a^4\ge2c^2a^2\)
\(\Rightarrow a^4+b^4+b^4+c^4+c^4+a^4\ge2a^2b^2+2b^2c^2+2c^2a^2\)
\(\Rightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)( 1 )
Ta lại có :
\(a^2b^2+b^2c^2\ge2ab^2c\)
\(b^2c^2+c^2a^2\ge2bc^2a\)
\(c^2a^2+a^2b^2\ge2ca^2b\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge ab^2c+bc^2a+ca^2b=abc\left(a+b+c\right)\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c\)( Đpcm )
Ta có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c>0\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-b^2ac-c^2ab\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2-2a^2c^2-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2-\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)\)\(+\left(b^2c^2+c^2a^2-2c^2ab\right)+\left(a^2b^2+c^2a^2-2a^2bc\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a,b,c
CMR:\(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d\right)\forall a,b,c,d,e\varepsilon R\)
Đề thiếu rồi nhé: \(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)
Quá ez:))
Ta có: \(a^2+b^2+c^2+d^2+e^2\)
\(=\left(\frac{a^2}{4}+b^2\right)+\left(\frac{a^2}{4}+c^2\right)+\left(\frac{a^2}{4}+d^2\right)+\left(\frac{a^2}{4}+e^2\right)\)
\(\ge2\sqrt{\frac{a^2}{4}\cdot b^2}+2\sqrt{\frac{a^2}{4}\cdot c^2}+2\sqrt{\frac{a^2}{4}\cdot d^2}+2\sqrt{\frac{a^2}{4}\cdot e^2}\)
\(=ab+ac+ad+ae=a\left(b+c+d+e\right)\)
Dấu "=" xảy ra khi: \(\frac{a}{2}=b=c=d=e\)
Sửa đề a2 + b2 + c2 + d2 + e2 ≥ a( b + c + d + e )
a2 + b2 + c2 + d2 + e2 ≥ a( b + c + d + e )
<=> a2 + b2 + c2 + d2 + e2 ≥ ab + ac + ad + ae
Nhân 4 vào từng vế
<=> 4( a2 + b2 + c2 + d2 + e2 ) ≥ 4( ab + ac + ad + ae )
<=> 4a2 + 4b2 + 4c2 + 4d2 + 4e2 ≥ 4ab + 4ac + 4ad + 4ae
<=> 4a2 + 4b2 + 4c2 + 4d2 + 4e2 - 4ab - 4ac - 4ad - 4ae ≥ 0
<=> ( a2 - 4ab + 4b2 ) + ( a2 - 4ac + 4c2 ) + ( a2 - 4ac + 4d2 ) + ( a2 - 4ae + 4e2 ) ≥ 0
<=> ( a - 2b )2 + ( a - 2c )2 + ( a - 2d )2 + ( a - 2e )2 ≥ 0 ( đúng )
Vậy bđt được chứng minh
Dấu "=" xảy ra <=> \(b=c=d=e=\frac{a}{2}\)
1. Có bao nhiêu \(m\in Z\) \(\in\left[-30;40\right]\) để bpt sau đúng \(\forall x\in R\)
\(a.\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)\ge m\)
b.\(b.\left(x^2-2x+4\right)\left(x^2+3x+4\right)\ge mx^2\)
2. Tìm m để pt
\(\left(m+3\right)x-2\sqrt{x^2-1}+m-3=0\) có nghiệm \(x\ge1\)
1.a.
\(\left(x+1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)\ge m\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-10\right)\ge m\)
Đặt \(x^2+3x-10=t\ge-\dfrac{49}{4}\)
\(\Rightarrow\left(t+2\right)t\ge m\Leftrightarrow t^2+2t\ge m\)
Xét \(f\left(t\right)=t^2+2t\) với \(t\ge-\dfrac{49}{4}\)
\(-\dfrac{b}{2a}=-1\) ; \(f\left(-1\right)=-1\) ; \(f\left(-\dfrac{49}{4}\right)=\dfrac{2009}{16}\)
\(\Rightarrow f\left(t\right)\ge-1\)
\(\Rightarrow\) BPT đúng với mọi x khi \(m\le-1\)
Có 30 giá trị nguyên của m
1b.
Với \(x=0\) BPT luôn đúng
Với \(x\ne0\) BPT tương đương:
\(\dfrac{\left(x^2-2x+4\right)\left(x^2+3x+4\right)}{x^2}\ge m\)
\(\Leftrightarrow\left(x+\dfrac{4}{x}-2\right)\left(x+\dfrac{4}{x}+3\right)\ge m\)
Đặt \(x+\dfrac{4}{x}-2=t\) \(\Rightarrow\left[{}\begin{matrix}t\ge2\\t\le-6\end{matrix}\right.\)
\(\Rightarrow t\left(t+5\right)\ge m\Leftrightarrow t^2+5t\ge m\)
Xét hàm \(f\left(t\right)=t^2+5t\) trên \(D=(-\infty;-6]\cup[2;+\infty)\)
\(-\dfrac{b}{2a}=-\dfrac{5}{2}\notin D\) ; \(f\left(-6\right)=6\) ; \(f\left(2\right)=14\)
\(\Rightarrow f\left(t\right)\ge6\)
\(\Rightarrow m\le6\)
Vậy có 37 giá trị nguyên của m thỏa mãn
2.
Xét với \(x\ge1\)
\(m\left(x+1\right)+3\left(x-1\right)-2\sqrt{x^2-1}=0\)
\(\Leftrightarrow m+3\left(\dfrac{x-1}{x+1}\right)-2\sqrt{\dfrac{x-1}{x+1}}=0\)
Đặt \(\sqrt{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)
\(\Rightarrow m+3t^2-2t=0\)
\(\Leftrightarrow3t^2-2t=-m\)
Xét hàm \(f\left(t\right)=3t^2-2t\) trên \(D=[0;1)\)
\(-\dfrac{b}{2a}=\dfrac{1}{3}\in D\) ; \(f\left(0\right)=0\) ; \(f\left(\dfrac{1}{3}\right)=-\dfrac{1}{3}\) ; \(f\left(1\right)=1\)
\(\Rightarrow-\dfrac{1}{3}\le f\left(t\right)< 1\)
\(\Rightarrow\) Pt có nghiệm khi \(-\dfrac{1}{3}\le-m< 1\)
\(\Leftrightarrow-1< m\le\dfrac{1}{3}\)