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a ) Ta có : \(\left(ab+1\right)^2\ge4ab\)
\(\Leftrightarrow a^2b^2+2ab+1-4ab\ge0\)
\(\Leftrightarrow\left(ab-1\right)^2\ge0\)
=> BĐT luôn đúng
Dấu " = " xảy ra \(\Leftrightarrow ab=1\)
b ) Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(ab+1.2\right)^2\le\left(a^2+1^2\right)\left(b^2+2^2\right)=\left(a^2+1\right)\left(b^2+4\right)\)
Dấu " = " xảy ra \(\Leftrightarrow2a=b\)
c ) Áp dụng BĐT Cô - si cho 2 số không âm , ta có :
\(4a^2+b^2\ge2\sqrt{4a^2.b^2}=4ab\)
\(\Rightarrow2\left(4a^2+b^2\right)\ge4a^2+4ab+b^2=\left(2a+b\right)^2\)
Dấu " = " xảy ra \(\Leftrightarrow2a=b\)
d ) \(x^5+y^5\ge xy\left(x^3+y^3\right)\)
\(\Leftrightarrow x^5-x^4y-y^4x+y^5\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\)
Vì x ; y > 0 => BĐT luôn đúng
Dấu " = " xảy ra \(\Leftrightarrow x=y\)
1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)
\(\ge\left(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\) \(\left(1\right)\)
Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\) ( Do abc=1 )
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)
Mà \(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\) (đpcm)
\(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4+a^3b+ab^3\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4-a^3b-ab^3\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right]\ge0\) * đúng *
b
Hiển hiên
\(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)
\(\Leftrightarrow a^4+b^4-a^3b-ab^3\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
Dấu "=" xảy ra <=> a=b
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2