Choa,b,c >0.
CMR: \(\dfrac{a^4}{b^2\left(c+a\right)}+\dfrac{b^4}{c^2\left(a+b\right)}+\dfrac{c^4}{a^2\left(b+c\right)}>=\dfrac{a+b+c}{2}\)
Cho a,b,c>0 thỏa mãn a+b+c=3 CMR:
\(\dfrac{a^4}{\left(a+2\right)\left(b+2\right)}+\dfrac{b^4}{\left(b+2\right)\left(c+2\right)}+\dfrac{c^4}{\left(c+2\right)\left(a+2\right)}\ge\dfrac{1}{3}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^4}{(a+2)(b+2)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\geq 4\sqrt[4]{\frac{a^4}{27.27.9}}=\frac{4a}{9}\)
\(\frac{b^4}{(b+2)(c+2)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\geq \frac{4b}{9}\)
\(\frac{c^4}{(c+2)(a+2)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\geq \frac{4c}{9}\)
Cộng theo vế và rút gọn:
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}+\frac{2(a+b+c)}{27}+\frac{7}{9}\geq\frac{4(a+b+c)}{9}\)
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}\geq \frac{10(a+b+c)}{27}-\frac{7}{9}=\frac{30}{27}-\frac{7}{9}=\frac{1}{3}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
cho a,b,c>0, CMR:
\(\left(a+b+\dfrac{1}{4}\right)^2+\left(b+c+\dfrac{1}{4}\right)^2+\left(c+a+\dfrac{1}{4}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)
\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)
Thật vậy, ta có:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)
Vậy ta cần chứng minh:
\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c
Cho a,b,c là các số khác 0 thỏa a+b+c=0.Cmr:
\(\dfrac{a^4}{a^4-\left(b^2-c^2\right)^2}+\dfrac{b^4}{b^4-\left(c^2-a^2\right)^2}+\dfrac{c^4}{c^4-\left(a^2-b^2\right)^2}=\dfrac{3}{4}\)
\(\frac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}=\frac{a^4}{\left[\left(a-b\right)\left(a+b\right)+c^2\right]\left[\left(a-c\right)\left(a+c\right)+b^2\right]}\)
\(\frac{a^4}{\left[-c\left(a-b\right)+c^2\right]\left[-b\left(a-c\right)+b^2\right]}=\frac{a^4}{4bc\left(b+c\right)^2}=\frac{a^4}{4a^2bc}\)
Tương tự với 2 phân thức còn lại, ta cũng có : \(\frac{b^4}{b^4-\left(c^2-a^2\right)^2}=\frac{b^4}{4ab^2c};\frac{c^4}{c^4-\left(a^2-b^2\right)^2}=\frac{c^4}{4abc^2}\)
\(VT=\frac{a^4}{4a^2bc}+\frac{b^4}{4ab^2c}+\frac{c^4}{4abc^2}=\frac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}=\frac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\)
\(VT=\frac{a^3+b^3+c^3}{4abc}\)
Mà \(a+b+c=0\) nên \(a^3+b^3+c^3=3abc\) ( tự cm )
\(\Rightarrow\)\(VT=\frac{3abc}{4abc}=\frac{3}{4}\) ( đpcm )
Chúc bạn học tốt ~
Cho a,b,c là các số khác 0 thỏa a+b+c=0.Cmr:
\(\dfrac{a^4}{a^4-\left(b^2-c^2\right)^2}+\dfrac{b^4}{b^4-\left(c^2-a^2\right)^2}+\dfrac{c^4}{c^4-\left(a^2-b^2\right)^2}=\dfrac{3}{4}\)
Đặt :
\(A=\)\(\dfrac{a^4}{a^4-\left(b^2-c^2\right)^2}+\dfrac{b^4}{b^4-\left(c^2-a^2\right)^2}+\dfrac{c^4}{c^4-\left(a^2-b^2\right)}\)
\(=\dfrac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}+\dfrac{b^4}{\left(b^2-c^2+a^2\right)\left(b^2+c^2-a^2\right)}+\dfrac{c^4}{\left(c^2-a^2+b^2\right)\left(c^2+a^2-b^2\right)}\)
Ta có : \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\)
Tương tự :
+) \(a^2-b^2+c^2=-2ac\)
+) \(b^2+c^2-a^2=-2bc\)
\(\Leftrightarrow A=\dfrac{a^4}{\left(-2ac\right)\left(-2ab\right)}+\dfrac{b^4}{\left(-2ab\right)\left(-2bc\right)}+\dfrac{c^4}{\left(-2bc\right)\left(-2ac\right)}\)
\(=\dfrac{a^4}{4a^2bc}+\dfrac{b^4}{4ab^2c}+\dfrac{c^4}{4abc^2}\)
\(=\dfrac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}\)
\(=\dfrac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\) (cậu tự chứng minh \(a^3+b^3+c^3=3abc\) nhé)
\(=\dfrac{3a^2b^2c^2}{4a^2b^2c^2}\)
\(=\dfrac{3}{4}\)
Vậy..
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
\(1,Cho.a,b,c\ge1.CMR:\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\ge\left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\)
2, Cho a,b,c>0.CMR:
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ac+b^2}+\dfrac{c+a}{ab+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
1.
BĐT cần chứng minh tương đương:
\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)
Ta có:
\(\left(ab-1\right)^2=a^2b^2-2ab+1=a^2b^2-a^2-b^2+1+a^2+b^2-2ab\)
\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)
Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)
\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)
Do \(a;b;c\ge1\) nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:
\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)
\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Câu 2 em kiểm tra lại đề có chính xác chưa
2.
Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS
Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)
\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)
BĐT cần chứng minh tương đương:
\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)
\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)
Đúng theo (1)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c là các số thực dương CMR : \(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)
\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\dfrac{9}{4}\)
\(\Rightarrow\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c >0 và abc=1. Tìm min:
\(P=\dfrac{a^4+b^4}{\left(a^2+b^2\right)\left(a+b\right)}+\dfrac{b^4+c^4}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{a^4+c^4}{\left(a^2+c^2\right)\left(a+c\right)}\)