Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xài Bunhiacopxki thì bài này sẽ hơi dài:
Đặt vế trái là P
Ta có:
\(\left(\dfrac{1}{4}+4\right)\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Leftrightarrow\dfrac{17}{4}\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)
\(\Rightarrow\sqrt{x^2+\dfrac{1}{x^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{2}{x}\right)\)
Tương tự:
\(\sqrt{y^2+\dfrac{1}{y^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{y}{2}+\dfrac{2}{y}\right)\) ; \(\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{z}{2}+\dfrac{2}{z}\right)\)
Cộng vế: \(P\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{y}{2}+\dfrac{z}{2}+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right)\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{36}{x+y+z}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{9}{4\left(x+y+z\right)}+\dfrac{135}{4\left(x+y+z\right)}\right)\)
\(P\ge\dfrac{1}{\sqrt{17}}\left(2\sqrt{\dfrac{9\left(x+y+z\right)}{4\left(x+y+z\right)}}+\dfrac{135}{4.\dfrac{3}{2}}\right)=\dfrac{3}{2}\sqrt{17}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Áp dụng BĐT Cosi ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)
Cmtt:
\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)
\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)
Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)
Tương tự cho 2 BĐT trên ta có:
\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)=\dfrac{1}{3}VT\)
Xảy ra khi \(x=y=z\)
Đặt \(\left(\dfrac{1}{\sqrt{x}};\dfrac{1}{\sqrt{y}};\dfrac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}=1\)
Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)
Thật vậy, ta có:
\(1=\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)
\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)
Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)