Cho a, b, c > 0 thoả mãn: \(a+b+c=3\). Chứng minh: \(\left(a+b+ab\right)\left(b+c+bc\right)\left(c+a+ca\right)\ge27abc\)
Cho 3 số thực dương a,b,c thoả mãn: a2+b2+c2=1 .Chứng minh:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}+\dfrac{b^5+c^5}{bc\left(b+c\right)}+\dfrac{c^5+a^5}{ca\left(c+a\right)}\ge3\left(ab+bc+ca\right)-2\)
Ta chứng minh BĐT sau cho các số dương:
\(x^5+y^5\ge xy\left(x^3+y^3\right)\)
\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
Áp dụng:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)
Tương tự và cộng lại:
\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)
\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)
\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)
Cho a, b, c là các số thực dương thoả mãn a+b+c=3. Chứng minh rằng:
\(\frac{a}{b^2\left(ca+1\right)}+\frac{b}{c^2\left(ab+1\right)}+\frac{c}{a^2\left(bc+1\right)}\ge\frac{9}{\left(1+abc\right)\left(ab+bc+ca\right)}\)
Theo bđt Cauchy - Schwart ta có:
\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)
\(=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)+3a^2b^2c^2}\)
Đặt \(ab+bc+ca=x;abc=y\).
Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)
\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )
Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1
làm sao mà \(x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\)lại luôn đúng
Cho 3 số thực dương a,b,c thoả mãn : ab+bc+ca=3 .Chứng minh :
\(\dfrac{1}{1+a^2\left(b+c\right)}+\dfrac{1}{1+b^2\left(c+a\right)}+\dfrac{1}{1+c^2\left(a+b\right)}\le\dfrac{1}{abc}\)
\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)
\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)
Cho a,b,c là các số thực dương thoả mãn \(ab+bc+ca=1\) . Chứng minh rằng:
\(\left(a^2+2b^2+3\right)\left(b^2+2c^2+3\right)\left(c^2+2a^2+3\right)\ge64\left(a^2+b^2+c^2\right)\)
Bất đẳng thức sai với [a = 35/256, b = 5/16, c = 3921/1840 ]
Helppppppppppppppppppp
Cho a,b,c là các số thực dương thoả mãn a +b + c <1 . Chứng minh rằng \(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+\left(a+b\right)}+\dfrac{1}{bc+\left(b+c\right)}+\dfrac{1}{ca+\left(c+a\right)}< \dfrac{87}{2}\)
Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).
cho a,b,c thoả mãn \(\frac{1}{bc-a^2}+\frac{1}{ca-b^2}+\frac{1}{ab-c^2}=0\)
chứng minh rằng \(y=\frac{a}{\left(bc-a^2\right)^2}+\frac{b}{\left(ac-b^2\right)^2}+\frac{c}{\left(ab-c^2\right)^2}=0\)
Cho a,b,c thực dương thoả mãn abc=1, chứng minh:
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\)
Cho a, b, c > 0 và a + b + c = 3. Chứng minh rằng \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(ab+c\right)\left(bc+a\right)\left(ca+b\right)\)
Cho a,b,c>0. Chứng minh
\(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
Ta có: \(a+b+c\ge3\sqrt[3]{abc}\)
\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)(1)
Ta có: \(\left(a-b\right)^3+\left(b-c\right)^2+\left(c-a\right)^3\)
\(=\left(a-b\right)^3+3\left(a-b\right)^2\left(b-c\right)+3\left(a-b\right)\left(b-c\right)^2+\left(b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)^2\left(b-c\right)-3\left(a-b\right)\left(b-c\right)^2\)
\(=\left(a-b+b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Ta có: \(a-b+b-c+c-a\ge3\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow9abc\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)(2)
Từ (1), (2) ta có: \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
Dấu "=" xảy ra khi \(a=b=c\)