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1) Áp dụng bất đẳng Bunyakovsky dạng cộng mẫu ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)
\(=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)
(Cauchy 3 số) Dấu "=" xảy ra khi: a = b = c
2) Áp dụng kết quả phần 1 ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^2+c^3\right)^2}{3\cdot\frac{1}{3}}=\left(a^3+b^3+c^3\right)^2\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2\) (do x+y=1)
\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)
Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)
Vậy \(x^3+y^3\ge\dfrac{1}{4}\)
2.
a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))
Đẳng thức xảy ra \(\Leftrightarrow a=b\)
b) Lần trước mk giải rồi nhá
3.
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)
b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)
\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )
Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
Ta có: \(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\) và \(\left\{{}\begin{matrix}a^2+1\ge2a\\b^2+1\ge2b\\c^2+1\ge2c\end{matrix}\right.\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ca\right)=12\)
\(\Rightarrow a^2+b^2+c^2\ge3\left(1\right)\)
Ta lại có:
\(\left\{{}\begin{matrix}\dfrac{a^3}{b}+ab\ge2a^2\\\dfrac{b^3}{c}+bc\ge2b^2\\\dfrac{c^3}{a}+ca\ge2c^2\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-ab-bc-ca\ge a^2+b^2+c^2\left(2\right)\)
Từ (1) và (2) \(\RightarrowĐPCM\)
Áp dụng bđt cô-si, ta có: \(a+b^2\le\dfrac{a^2+1}{2}+b^2=\dfrac{a^2+2b^2+1}{2}\)
=>\(\dfrac{2a^2}{a+b^2}\ge\dfrac{4a^2}{a^2+2b^2+1}\)
CMTT: Khi đó: \(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{4a^2}{a^2+2b^2+1}+\dfrac{4b^2}{b^2+2c^2+1}+\dfrac{4c^2}{c^2+2a^2+1}\)
Áp dụng bđt Sơ-vác, ta có:
\(\dfrac{4a^4}{a^4+2a^2b^2+a^2}+\dfrac{4b^4}{b^4+2b^2c^2+b^2}+\dfrac{4c^4}{c^4+2c^2a^2+c^2}\ge\dfrac{4\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=\dfrac{4.3^2}{3^2+3}=3\)
Do đó: \(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge\dfrac{4a^2}{a^2+2b^2+1}+\dfrac{4b^2}{b^2+2c^2+1}+\dfrac{4c^2}{c^2+2a^2+1}\ge3\)
Vì \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)
=>\(\dfrac{2a^2}{a+b^2}+\dfrac{2b^2}{b+c^2}+\dfrac{2c^2}{c+a^2}\ge a+b+c\)
Dấu "=" xảy ra khi a=b=c=1
=>ĐPCM
Ko lq nhưng ta chuẩn hóa \(a+b+c=3\). So:
\(M\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
\(A=\dfrac{a^4}{a\left(b+c\right)}+\dfrac{b^4}{b\left(a+c\right)}+\dfrac{c^4}{c\left(a+b\right)}\)
\(\Rightarrow A\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2ab+2ac+2bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+a^2+c^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{2}=1\)
\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}a=b=c=\dfrac{\sqrt{6}}{3}\\a=b=c=\dfrac{-\sqrt{6}}{3}\end{matrix}\right.\)
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