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1.
\(\left(x+y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{1}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{1}{9}\right)\left(4x^2+9y^2\right)=\dfrac{169}{36}\)
\(\Rightarrow-\dfrac{13}{6}\le x+y\le\dfrac{13}{6}\)
Dấu "=" lần lượt xảy ra tại \(\left(-\dfrac{3}{2};-\dfrac{2}{3}\right)\) và \(\left(\dfrac{3}{2};\dfrac{2}{3}\right)\)
2.
\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y+\left(-\dfrac{1}{3}\right).6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)
\(\Rightarrow\left|y-2x\right|\le\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\mp\dfrac{2}{5};\pm\dfrac{9}{20}\right)\)
3.
\(B^2=\left(6.\sqrt{x-1}+8\sqrt{3-x}\right)^2\le\left(6^2+8^2\right)\left(x-1+3-x\right)=200\)
\(\Rightarrow B\le2\sqrt{10}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{x-1}}{6}=\dfrac{\sqrt{3-x}}{8}\Leftrightarrow x=\dfrac{43}{25}\)
\(B=6\sqrt{x-1}+6\sqrt{3-x}+2\sqrt{3-x}\ge6\sqrt{x-1}+6\sqrt{3-x}\)
\(B\ge6\left(\sqrt{x-1}+\sqrt{3-x}\right)\ge6\sqrt{x-1+3-x}=6\sqrt{2}\)
\(B_{min}=6\sqrt{2}\) khi \(\sqrt{3-x}=0\Rightarrow x=3\)
4.
\(49=\left(3a+4b\right)^2=\left(\sqrt{3}.\sqrt{3}a+2.2b\right)^2\le\left(3+4\right)\left(3a^2+4b^2\right)\)
\(\Rightarrow3a^2+4b^2\ge\dfrac{49}{7}=7\)
Dấu "=" xảy ra khi \(a=b=1\)
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5.
\(\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+\sqrt{ac}+\sqrt{ab}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tương tự:
\(\dfrac{b}{b+\sqrt{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
\(\dfrac{c}{c+\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Cộng vế với vế:
\(P\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
Dấu "=" xảy ra khi \(a=b=c\)
6.
\(P=\dfrac{a}{1+b-a}+\dfrac{b}{1+c-b}+\dfrac{c}{1+a-c}\)
Thay \(1=a+b+c\)
\(\Rightarrow P=\dfrac{a}{2b+c}+\dfrac{b}{2c+a}+\dfrac{c}{2a+b}\)
\(P=\dfrac{a^2}{2ab+ac}+\dfrac{b^2}{2bc+ab}+\dfrac{c^2}{2ac+bc}\)
\(P\ge\dfrac{\left(a+b+c\right)^2}{3ab+3bc+3ca}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
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Thầy Nguyễn Việt Lâm ơi giúp em mấy bài này với.Em sắp phải nộp rồi ạ - Hoc24
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4.
\(ab+bc+ca=3abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)
\(S=\sum\dfrac{\dfrac{1}{y^2}}{\dfrac{1}{x}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}=\sum\dfrac{x^3}{x^2+y^2}=\sum\left(x-\dfrac{xy^2}{x^2+y^2}\right)\)
\(S\ge\sum\left(x-\dfrac{xy^2}{2xy}\right)=\sum\left(x-\dfrac{y}{2}\right)=\dfrac{x+y+z}{2}=\dfrac{3}{2}\)
\(S_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)
5.
Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)
Đặt vế trái là P
\(P=\dfrac{z^3}{x^2+z^2}+\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}\)
Quay lại dòng 3 của bài số 4