Not anonymous
- 143
- 58
Define ##x = (a + b)## and ##y = (c + d)##. Then the LHS of the inequality in the question, i.e. $$
\dfrac {1} {\dfrac {1} {a} + \dfrac {1} {b}} + \dfrac {1} {\dfrac {1} {c} + \dfrac {1} {d}}
$$
is equivalent to $$
\dfrac {ab} {a+b} + \dfrac {cd} {c+d} = \dfrac {ab} {x} + \dfrac {cd} {y} = \dfrac {aby + cdx} {xy}
$$
Similarly, the RHS can be expressed in terms of ##x## and ##y## as
##\dfrac {1} {\dfrac {1} {a+c} + \dfrac {1} {b+d}} = \dfrac {(a+c)(b+d)} {a+b+c+d} = \dfrac {(a+c)(b+d)} {x+y}##
##\Rightarrow LHS - RHS = \dfrac{aby+cdx} {xy} - \dfrac{(a+c)(b+d)} {x+y} =##
##\dfrac{(abxy + aby^{2} +cdx^{2} + cdxy) - (abxy + adxy + bcxy + cdxy)} {xy(x+y)} =##
##\dfrac {aby^{2} + cdx^{2} - adxy - bcxy} {xy(x+y)} = \dfrac {(ay - cx)(by - dx)} {xy(x+y)} =##
##\dfrac {(ac + ad - ca - cb)(bc + bd - da - db)} {xy(x+y)} =##
##\dfrac {(ad - bc)(bc - ad)} {xy(x+y)} = \dfrac {-(ad - bc)^{2}} {xy(x+y)}##
The denominator of the final expression ##xy(x+y)## must be positive since the ##a, b, c, d## are all positive and therefore ##x, y## too are positive. And ##-(ad - bc)^{2} \leq 0## for any real-valued ##a, b, c, d##. Hence ##LHS - RHS \leq {0} \Rightarrow \text{LHS} \leq \text{RHS}##, hence proving the inequality.
\dfrac {1} {\dfrac {1} {a} + \dfrac {1} {b}} + \dfrac {1} {\dfrac {1} {c} + \dfrac {1} {d}}
$$
is equivalent to $$
\dfrac {ab} {a+b} + \dfrac {cd} {c+d} = \dfrac {ab} {x} + \dfrac {cd} {y} = \dfrac {aby + cdx} {xy}
$$
Similarly, the RHS can be expressed in terms of ##x## and ##y## as
##\dfrac {1} {\dfrac {1} {a+c} + \dfrac {1} {b+d}} = \dfrac {(a+c)(b+d)} {a+b+c+d} = \dfrac {(a+c)(b+d)} {x+y}##
##\Rightarrow LHS - RHS = \dfrac{aby+cdx} {xy} - \dfrac{(a+c)(b+d)} {x+y} =##
##\dfrac{(abxy + aby^{2} +cdx^{2} + cdxy) - (abxy + adxy + bcxy + cdxy)} {xy(x+y)} =##
##\dfrac {aby^{2} + cdx^{2} - adxy - bcxy} {xy(x+y)} = \dfrac {(ay - cx)(by - dx)} {xy(x+y)} =##
##\dfrac {(ac + ad - ca - cb)(bc + bd - da - db)} {xy(x+y)} =##
##\dfrac {(ad - bc)(bc - ad)} {xy(x+y)} = \dfrac {-(ad - bc)^{2}} {xy(x+y)}##
The denominator of the final expression ##xy(x+y)## must be positive since the ##a, b, c, d## are all positive and therefore ##x, y## too are positive. And ##-(ad - bc)^{2} \leq 0## for any real-valued ##a, b, c, d##. Hence ##LHS - RHS \leq {0} \Rightarrow \text{LHS} \leq \text{RHS}##, hence proving the inequality.