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Áp dụng BĐT Cô si dạng phân số ta có :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> ĐPCM .
b) Vì a,b,c > 0 .
Áp dụng BĐT Cô si ta có :
\(\dfrac{a^2}{b}+b\ge2a\) (1)
Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2b\) (2)
\(\dfrac{c^2}{a}+a\ge2c\) (3)
Cộng từng vế => ĐPCM .
Xét:
\(\dfrac{a^2}{b^2+c^2}-\dfrac{a}{b+c}=\dfrac{a\left(ab+ac-b^2-c^2\right)}{\left(b^2+c^2\right)\left(b+c\right)}=\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\left(1\right)\)
Tương tự:
\(\dfrac{b^2}{c^2+a^2}-\dfrac{b}{c+a}=\dfrac{bc\left(b-c\right)+ba\left(b-a\right)}{\left(c^2+a^2\right)\left(c+a\right)}\) (2)
\(\dfrac{c^2}{a^2+b^2}-\dfrac{c}{a+b}=\dfrac{ca\left(c-a\right)+cb\left(c-b\right)}{\left(a^2+b^2\right)\left(a+b\right)}\) (3)
Cộng từng vế (1)(2)(3) ta được:
\(\left(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\right)-\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]+ac\left(a-c\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]+bc\left(b-c\right)\left[\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\) => ĐPCM
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 \(a=b=c\)
Áp dụng BĐT Cô - Si , ta có :
\(\dfrac{a}{b^2}+\dfrac{1}{a}\) ≥ \(2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=2.\dfrac{1}{b}\left(a,b>0\right)\left(1\right)\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\text{ ≥ }2\sqrt{\dfrac{b}{c^2}.\dfrac{1}{b}}=2.\dfrac{1}{c}\left(b,c>0\right)\left(2\right)\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\text{≥}2\sqrt{\dfrac{c}{a^2}.\dfrac{1}{c}}=2.\dfrac{1}{a}\left(a,c>0\right)\left(3\right)\)
Từ ( 1 ; 2 ; 3) Ta có :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
⇔\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\) ≥ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Ta có:
\(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2}\)
\(=a+b+c-\dfrac{ab^2}{1+b^2}-\dfrac{bc^2}{1+c^2}-\dfrac{ca^2}{1+a^2}\)
\(\ge3-\dfrac{ab^2}{2b}-\dfrac{bc^2}{2c}-\dfrac{ca^2}{2a}\)
\(=3-\dfrac{1}{2}\left(ab+bc+ca\right)\ge3-\dfrac{1}{2}.\dfrac{\left(a+b+c\right)^2}{3}\)
\(=3-\dfrac{3}{2}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=1\)
Đặt P=\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Không mất tính tổng quát giả sử a ≥b ≥ c , thế thì \(\dfrac{1}{b+c}\ge\dfrac{1}{c+a}\ge\dfrac{1}{a+b}\) .Áp dụng bất đẳng thức Chebyshev cho hai dãy đơn điệu cùng chiều ta có :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{1}{3}\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\left(\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\right)\)
Hay \(P\ge\dfrac{1}{3}\left(P+3\right)\) nghĩa là \(P\ge\dfrac{3}{2}^{\left(đpcm\right)}\)
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)
\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)
\(\ge\dfrac{3abc}{2abc}+\dfrac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}+\dfrac{2\left(b^2+c^2\right)}{2a^2+b^2+c^2}+\dfrac{2\left(a^2+c^2\right)}{2b^2+a^2+c^2}\)
\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)
Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:
\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)
\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)
\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế: \(VT+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
ko xoắn 1 dòng thôi
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
mình cảm ơn nha