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\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Ta có
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)
\(=\dfrac{abc}{ab+b+abc}+\dfrac{abc}{bc+c+abc}+\dfrac{1}{ca+a+1}\)
\(=\dfrac{abc}{b\left(ac+a+1\right)}+\dfrac{abc}{c\left(ab+b+1\right)}+\dfrac{1}{ac+a+1}\)
\(=\dfrac{ac}{ac+a+1}+\dfrac{ab}{ab+b+1}+\dfrac{1}{ac+a+1}\)
\(=\dfrac{ac+1}{ac+a+1}+\dfrac{ab}{ab+b+abc}\)
\(=\dfrac{ac+1}{ac+a+1}+\dfrac{ab}{b\left(ac+a+1\right)}=\dfrac{ac+a+1}{ac+a+1}=1\) (đpcm)
Ta có: \(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{b}{abc+ab+b}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{abc}{c\left(ab+b+1\right)}+\dfrac{b}{ab+b+1}\)
=\(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab+b+1}+\dfrac{b}{ab+b+1}\)
=\(\dfrac{ab+b+1}{ab+b+1}\)=1
Suy ra:
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)=1(abc=1)
(đpcm)
Đề đúng đây nhé
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
Áp dụng BĐT Cosi ta có:
\(a^2+bc\ge2a\sqrt{bc}\)
\(\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2a\sqrt{bc}}\)
Cmtt: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{2b\sqrt{ac}}\)
\(\dfrac{1}{c^2+ab}\le\dfrac{1}{2c\sqrt{ab}}\)
Cộng vế theo vế ta được
\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2a\sqrt{bc}}+\dfrac{1}{2b\sqrt{ac}}+\dfrac{1}{2c\sqrt{ab}}\)
\(\Leftrightarrow\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\)
Mà \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\) (C/m sau)
Nên \(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
Chứng minh \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)
\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)
\(\text{}\Leftrightarrow\text{}\text{}2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\le2a+2b+2c\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\left(lđ\right)\)
Lời giải:
a)
Sử dụng pp biến đổi tương đương:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\Leftrightarrow \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)
\(\Leftrightarrow (ab+1)(a^2+b^2+2)\geq 2(a^2b^2+a^2+b^2+1)\)
\(\Leftrightarrow ab(a^2+b^2)+2ab\geq 2a^2b^2+a^2+b^2\)
\(\Leftrightarrow ab(a^2+b^2-2ab)-(a^2+b^2-2ab)\geq 0\)
\(\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0\)
\(\Leftrightarrow (ab-1)(a-b)^2\geq 0\) (luôn đúng với mọi $ab\geq 1$)
Ta có đpcm.
b) Áp dụng công thức của phần a ta có:
\(\frac{1}{a^4+1}+\frac{1}{b^4+1}\geq \frac{2}{1+(ab)^2}\)
Tiếp tục áp dụng công thức phần a: \(\frac{1}{1+(ab)^2}+\frac{1}{1+b^4}\geq \frac{2}{1+ab^3}\)
Do đó:
\(\frac{1}{a^4+1}+\frac{3}{b^4+1}\geq \frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{b^4+1}+\frac{3}{c^4+1}\geq \frac{4}{1+bc^3}; \frac{1}{c^4+1}+\frac{3}{a^4+1}\geq \frac{4}{1+ca^3}\)
Cộng theo vế các BĐT trên thu được:
\(4\left(\frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\right)\geq 4\left(\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\right)\)
\(\Leftrightarrow \frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\geq \frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c=1$
\(\dfrac{1}{1+a+ab}+\dfrac{a}{a+ab+abc}+\dfrac{1}{abc+ac+c}\)
\(=\dfrac{1}{1+a+ab}+\dfrac{a}{1+a+ab}+\dfrac{1}{c\left(1+a+ab\right)}\)
\(=\dfrac{ac+c+1}{c\left(1+a+ab\right)}=\dfrac{c\left(a+1+ab\right)}{c\left(1+a+ab\right)}=1\)