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Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=z\\\sqrt{b^2+c^2}=x\\\sqrt{c^2+a^2}=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=z^2\\b^2+c^2=x^2\\c^2+a^2=y^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
Điều kiện đề bài thành: \(x+y+z=3\sqrt{2}\)
Ta có:
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
\(\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
\(=\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\)
\(=\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-x-y-z\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-x-y-z\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2\sqrt{2}}.3\sqrt{2}=\dfrac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{2}\) hay \(a=b=c=1\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Ko lq nhưng ta chuẩn hóa \(a+b+c=3\). So:
\(M\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
Đây là bài IMO 2001 và không cần điều kiện \(a+b+c=1\)
Áp dụng Holder:
\(P.P.\left[a\left(a^2+8bc\right)+b\left(b^2+8ac\right)+c\left(c^2+8ab\right)\right]\ge\left(a+b+c\right)^3\)
\(\Leftrightarrow P^2\ge\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}=\dfrac{a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a^3+b^3+c^3+24abc}\)
\(\Rightarrow P^2\ge\dfrac{a^3+b^3+c^3+3.2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{a^3+b^3+c^3+24abc}=1\)
\(\Rightarrow P\ge1\)
Bài 1:
Áp dụng BĐT Bunhiacopxky:
\(M^2=(a\sqrt{9b(a+8b)}+b\sqrt{9a(b+8a)})^2\)
\(\leq (a^2+b^2)(9ab+72b^2+9ab+72a^2)\)
\(\Leftrightarrow M^2\leq (a^2+b^2)(72a^2+72b^2+18ab)\)
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 18ab\leq 9(a^2+b^2)\)
Do đó, \(M^2\leq (a^2+b^2)(72a^2+72b^2+9a^2+9b^2)=81(a^2+b^2)^2\)
\(\Leftrightarrow M\leq 9(a^2+b^2)\leq 144\)
Vậy \(M_{\max}=144\Leftrightarrow a=b=\sqrt{8}\)
Bài 6:
\(a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\)
Vì \(a>1\rightarrow a-1>0\). Do đó áp dụng BĐT Am-Gm cho số dương\(a-1,\frac{1}{a-1}\) ta có:
\((a-1)+\frac{1}{a-1}\geq 2\sqrt{\frac{a-1}{a-1}}=2\)
\(\Rightarrow a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\geq 3\) (đpcm)
Dấu bằng xảy ra khi \(a-1=1\Leftrightarrow a=2\)
Bài 3:
Xét \(\sqrt{a^2+1}\). Vì \(ab+bc+ac=1\) nên:
\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
\(\Rightarrow \sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM có: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)
hay \(\sqrt{a^2+1}\leq \frac{2a+b+c}{2}\)
Hoàn toàn tương tự với các biểu thức còn lại và cộng theo vế:
\(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\leq \frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}=2(a+b+c)\)
Ta có đpcm. Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bài 4:
Ta có:
\(A=\frac{8a^2+b}{4a}+b^2=2a+\frac{b}{4a}+b^2\)
\(\Leftrightarrow A+\frac{1}{4}=2a+\frac{b+a}{4a}+b^2=2a+b+\frac{b+a}{4a}+b^2-b\)
Vì \(a+b\geq 1, a>0\) nên \(A+\frac{1}{4}\geq a+1+\frac{1}{4a}+b^2-b\)
Áp dụng BĐT AM-GM:
\(a+\frac{1}{4a}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\Rightarrow A+\frac{1}{4}\geq 2+b^2-b=\left(b-\frac{1}{2}\right)^2+\frac{7}{4}\geq \frac{7}{4}\)
\(\Leftrightarrow A\geq \frac{3}{2}\).
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=\frac{1}{2}\)
Ta có BĐT \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\ge0\) (đúng)
\(\Rightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=1\)
Khi đó áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Tương tự cho 2 BĐT còn lại:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}=VP\)
Xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\)
Áp dụng BĐT Bu-nhi-a ta có:
\(\sqrt{a^2+1}=\sqrt{a^2+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{4\left(a^2+\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\right)}\)
\(\ge\dfrac{1}{2}\sqrt{\left(a+\dfrac{1}{\sqrt{3}}.3\right)^2}=\dfrac{1}{2}\sqrt{\left(a+\sqrt{3}\right)^2}=\dfrac{a+\sqrt{3}}{2}\left(a>0\right)\)
Tương tự ta cũng có: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{2b}{b+\sqrt{3}}\)
\(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{2c}{c+\sqrt{3}}\)
=> \(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)
\(\le2\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)\) (1)
Áp dụng BĐT phụ: \(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{x+y}\) ta có:
\(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\)
\(=\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}\)
\(\le\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{a+c}{a+c}+\dfrac{b+a}{a+b}+\dfrac{c+b}{b+c}\right)=\dfrac{3}{4}\) (2)
Từ (1); (2)
=> \(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le2.\dfrac{3}{4}=\dfrac{3}{2}\left(đpcm\right)\)
Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Chắc đề bị nhầm rồi.
\(\dfrac{a}{\sqrt{b+1}}+\dfrac{b}{\sqrt{c+1}}+\dfrac{c}{\sqrt{a+1}}\ge2\sqrt{2}\left(\dfrac{a}{3+b}+\dfrac{b}{3+c}+\dfrac{c}{3+a}\right)\)
\(\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(ab+bc+ca\right)}\ge2\sqrt{2}.\dfrac{9}{9+\dfrac{\left(a+b+c\right)^2}{3}}=2\sqrt{2}.\dfrac{9}{12}=\dfrac{3}{\sqrt{2}}\)
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
\(M\ge\dfrac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{c}+\sqrt{a}\right)^2}}{2}\)
\(M\ge\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)