Cho a+b+c=abc CMR:
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)
Cho a+b+c=abc CMR:
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)
Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
cho a,b,c là các số thực dương. Cmr
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)+\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Cho a,b,c không âm . CMR
\(2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(abc+1\right)\)
Cho a;b;c\(\in\)R .CMR \(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Cho a,b,c là các số thực dương.
\(CMR:\left(a^2+b\right)\left(b^2+c\right)\left(c^2+a\right)\ge abc\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
Nhân tung tóe + rút gọn ta được: \(\Sigma_{cyc}a^3b^2+\Sigma_{cyc}ab^3\ge abc\left(ab+bc+ca+a+b+c\right)\)
\(\Leftrightarrow\)\(\Sigma\frac{a^2b}{c}+\Sigma\frac{a^2}{b}\ge ab+bc+ca+a+b+c\) (*)
(*) đúng do \(\hept{\begin{cases}\frac{a^2b}{c}+bc\ge2ab\\\frac{a^2}{b}+b\ge2a\end{cases}}\Rightarrow\hept{\begin{cases}\Sigma\frac{a^2b}{c}\ge ab+bc+ca\\\Sigma\frac{a^2}{b}\ge a+b+c\end{cases}}\)
"=" \(\Leftrightarrow\)\(a=b=c\)
Cho a,b,c dương thỏa mãn \(abc\ge1\)CMR
\(27\left(a^3+a^2+a+1\right)\left(b^3+b^2+b+1\right)\left(c^3+c^2+c+1\right)\ge64\left(a^2+a+1\right)\left(b^2+b+1\right)\left(c^2+c+1\right)\)
Chứng minh cái này đi: \(\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{2}{3}a+\frac{2}{3}\) ( gợi ý: bđt \(\Leftrightarrow\)\(\left(a-1\right)^2\left(a+1\right)\ge0\))
Tương tự với 2 ẩn kia \(\Rightarrow\)\(\Sigma\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{8}{27}\Pi\left(a+1\right)\ge\frac{64}{27}\sqrt{abc}\ge\frac{64}{27}\)
dấu "=" xảy ra khi \(a=b=c=1\)
cho a,b,c.>0 thoả mãn ab+bc+ac=1. CMR
\(\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2+\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2\ge8\sqrt{3}abc\)
\(1,Cho.a,b,c\ge1.CMR:\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\ge\left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\)
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 \(a=b=c\)